3. POPULATION GENETICS.

HARDY-WEINBERG LAW

population- this is a collection of individuals of the same species, occupying a certain area for a long time, freely interbreeding among themselves and relatively isolated from other individuals of the species.

The main pattern that allows you to explore genetic structure big populations, was established in 1908 independently by the English mathematician G. Hardy and the German physician W. Weinberg.

Hardy-Weinberg law: in an ideal population, the ratio of frequencies of genes and genotypes–constant from generation to generation.

signs ideal population: population size big, exists panmixia(no restrictions to the free choice of a partner), no mutations on this basis, does not work natural selection, absent inflow And outflow of genes.

First position The Hardy-Weinberg law states: sum of allele frequencies one gene in a given population equal to one. This is written as follows:

p+ q = 1 ,

Where p– dominant allele frequency A,q- recessive allele frequency A. Both quantities are usually expressed in fractions of a unit, less often in percentages (then p+q = 100 %).

Second position Hardy-Weinberg law: sum of genotype frequencies one gene per population equal to one. The formula for calculating genotype frequencies is as follows:

p 2 + 2 pq + q 2 = 1 ,

Where p 2– the frequency of homozygous individuals for the dominant allele (genotype AA), 2pq- frequency of heterozygotes (genotype Aa), q 2 – the frequency of homozygous individuals for the recessive allele (genotype aa).

The derivation of this formula is: equilibrium population female and male individuals have the same frequencies as the A allele ( p), and allele a ( q). As a result of crossing female gametes ♀( p+q) with male ♂( p+q) and genotype frequencies are determined: ( p+q) (p+q) = p 2 + 2pq +q 2 .

Third position law: in an equilibrium population allele frequencies And genotype frequencies are preserved in a number of generations.

TASKS

3.1. In a population obeying the Hardy-Weinberg law, allele frequencies A And A respectively, are 0.8 and 0.2. Determine the frequencies of homozygotes and heterozygotes for these genes in the first generation.

Solution. Genotype frequencies are calculated using the Hardy-Weinberg equation:

p 2 + 2pq +q 2 = 1,

Where p is the frequency of the dominant gene, and q is the frequency of the recessive gene.

In this problem, the allele frequency A is 0.8, and the allele frequency A equals 0.2. Substituting these numerical values into the Hardy-Weinberg equation, we get the following expression:

0.82 + 2 × 0.8 × 0.2 + 0.22 = 1 or 0.64 + 0.32 + 0.04 = 1

It follows from the equation that 0.64 is the frequency of the dominant homozygous genotype ( AA), and 0.04 is the frequency of the recessive homozygous genotype ( aa). 0.32 – frequency of heterozygous genotype ( Ah).

3.2. In a fox population, there are 10 white individuals per 1000 red foxes. Determine percentage red homozygous, red heterozygous and white foxes in this population.

Solution.

According to the equation:

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Thus, there are 81% of red homozygous foxes in the population, 18% of red heterozygous foxes, and 1% of white foxes.

3.3. The brown-eyed allele is dominant over blue-eyed. In a population, both alleles occur with equal probability.

Father and mother are brown-eyed. What is the probability that their child will be blue-eyed?

Solution. If both alleles are equally common in a population, then it has 1/4 dominant homozygotes, 1/2 heterozygotes (both brown-eyed) and 1/4 recessive homozygotes (blue-eyed). Thus, if a person is brown-eyed, then two against one, that this is a heterozygote. So, the probability of being heterozygous is 2/3. The probability of passing the blue-eyed allele to offspring is 0 if the organism is homozygous, and 1/2 if it is heterozygous. Full Probability the fact that a given brown-eyed parent will pass on the blue eye allele to the offspring is 2/3x1/2, i.e. 1/3. In order for a child to be blue-eyed, it must receive an allele for blue eyes from each parent. This will happen with a probability of 1/3x1/3=1/9.

3.4. Cystic fibrosis of the pancreas affects individuals with a recessive homozygous phenotype and occurs in the population with a frequency of 1 in 2000. Calculate the carrier frequency.

Solution. Carriers are heterozygotes. Genotype frequencies are calculated using the Hardy-Weinberg equation:

p 2 + 2pq +q 2 = 1,

Where p 2 - frequency of dominant homozygous genotype, 2 pq is the frequency of the heterozygous genotype and q 2 - the frequency of the recessive homozygous genotype.

Cystic fibrosis of the pancreas affects individuals with a recessive homozygous phenotype; hence, q 2 = 1 in 2000, or 1/2000 = 0.0005. From here

Because the, p+q = 1, p= 1 – q = 1 – 0,0224 = 0,9776.

Thus, the frequency of the heterozygous phenotype (2 pq) \u003d 2 × (0.9776) × (0.0224) \u003d 0.044 \u003d 1 in 23 » 5%, i.e., carriers of the recessive gene for pancreatic cystic fibrosis make up about 5% of the population.

3.5. When examining the population of the city N (100,000 people), 5 albinos were found. Determine the frequency of occurrence of heterozygous carriers of the albinism gene.

Solution. Since albinos are recessive homozygotes ( aa), then, according to the Hardy-Weinberg law: recessive gene frequency, p+q = 1, from here, p= 1 – q; The frequency of heterozygotes is 2 pq.

Therefore, every 70th inhabitant of city N is a heterozygous carrier of the albinism gene.

3.6. In a population of 5,000 people, 64% are able to roll their tongue into a tube (dominant gene R), and 36% do not have this ability (recessive gene r). Calculate Gene Frequencies R And r and genotypes RR, Rr And rr in the population.

Solution. The frequency of occurrence of persons with genotypes RR And Rr equal to 0.64, and homozygotes rr(q 2) = 0.36. Based on this, the gene frequency r is equal to . And since p+q= 1, then p = 1 – q= 0.4, i.e. allele frequency R(p) is 0.4. If p= 0.4, then p 2 = 0.16. This means that the frequency of occurrence of individuals with the genotype RR is 16%.

So, the frequency of occurrence of genes R And r 0.4 and 0.6. Genotype Frequencies RR, Rr And rr are, respectively, 0.16, 0.48 and 0.36.

3.7. There are three albinism genotypes in the population A in ratio: 9/16 AA, 6/16 Ah and 1/16 aa. Is this population in a state of genetic equilibrium?

Solution. It is known that the population consists of 9/16 AA, 6/16 Ah and 1/16 aa genotypes. Does such a ratio correspond to the equilibrium in the population, expressed by the Hardy-Weinberg formula?

After converting the numbers, it becomes clear that the population according to the given trait is in equilibrium: (3/4)2 AA: 2×3/4×1/4 Ah: (1/4)2 aa. From here

3.8. Diabetes mellitus occurs in the population with a frequency of 1 in 200. Calculate the frequency of carrier heterozygotes.

3.9. Sickle cell anemia occurs in the human population with a frequency of 1: 700. Calculate the frequency of heterozygotes.

3.10. Share of individuals aa in a large population is 0.49. What fraction of the population is heterozygous for the gene A?

3.11. In the Drosophila population, the allele frequency b(black body color) is 0.1. Set the frequency of gray and black flies in the population and the number of homozygous and heterozygous individuals.

3.12. Does the following ratio of homozygotes and heterozygotes in the population correspond to the Hardy-Weinberg formula: 4096 AA : 4608 Ah : 1296 aa?

3.13. In one population, 70% of people are able to taste the bitter taste of phenylthiourea (PTU), and 30% do not distinguish its taste. The ability to taste FTM is determined by a dominant gene T. Determine the allele frequency T And t and genotypes TT, Tt And tt in this population.

3.14. Share of individuals AA in a large panmictic population is 0.09. What fraction of the population is heterozygous for the gene A?

3.15. Albinism in rye is inherited as an autosomal recessive trait. There are 84,000 plants in the surveyed area. Among them, 210 albinos were found.

Determine the frequency of the albinism gene in rye.

3.16.* In shorthorn cattle, the red color does not completely dominate the white color. Hybrids from crossing red with white have a roan suit. There are 4,169 reds, 3,780 roans and 756 whites registered in the shorthorn area.

Determine the frequency of genes for red and white color of cattle in the area.

3.17.* A single grain of wheat, heterozygous for a certain gene, accidentally fell on a deserted island A. It sprang up and gave rise to a series of generations that reproduced by self-pollination. What will be the proportion of heterozygous plants among representatives of the second, third, fourth, ..., nth generations, if the trait controlled by the gene under consideration does not affect the survival of plants and their ability to produce offspring under these conditions?

3.18.* Snyder examined 3643 people for the ability to taste phenylthiourea and found that 70.2% of them are "tasting" and 29.8% are "not tasting" this taste.

(a) What is the proportion of “non-sentient” children in “sentient” to “sentient” marriages?

b) What is the proportion of "not tasting" children in marriages "tasting" and "not tasting" phenylthiourea?

For genetic research, a person is an inconvenient object, since in a person: experimental crossing is impossible; a large number of chromosomes; puberty comes late; a small number of descendants in each family; equalization of living conditions for offspring is impossible.

A number of research methods are used in human genetics.

genealogical method

The use of this method is possible in the case when direct relatives are known - the ancestors of the owner of the hereditary trait ( proband) on the maternal and paternal lines in a number of generations or the descendants of the proband also in several generations. When compiling pedigrees in genetics, a certain system of notation is used. After compiling the pedigree, its analysis is carried out in order to establish the nature of the inheritance of the trait under study.

Conventions adopted in the preparation of pedigrees:
1 - man; 2 - woman; 3 - gender not clear; 4 - the owner of the studied trait; 5 - heterozygous carrier of the studied recessive gene; 6 - marriage; 7 - marriage of a man with two women; 8 - related marriage; 9 - parents, children and the order of their birth; 10 - dizygotic twins; 11 - monozygotic twins.

Thanks to the genealogical method, the types of inheritance of many traits in humans have been determined. So, according to the autosomal dominant type, polydactyly (an increased number of fingers), the ability to roll the tongue into a tube, brachydactyly (short fingers due to the absence of two phalanges on the fingers), freckles, early baldness, fused fingers, cleft lip, cleft palate, cataracts of the eyes, fragility of bones and many others. Albinism, red hair, susceptibility to polio, diabetes mellitus, congenital deafness, and other traits are inherited as autosomal recessive.

The dominant trait is the ability to roll the tongue into a tube (1) and its recessive allele is the absence of this ability (2).
3 - pedigree for polydactyly (autosomal dominant inheritance).

A number of traits are inherited sex-linked: X-linked inheritance - hemophilia, color blindness; Y-linked - hypertrichosis of the edge of the auricle, webbed toes. There are a number of genes located in homologous regions of the X and Y chromosomes, such as general color blindness.

The use of the genealogical method showed that in a related marriage, compared with an unrelated one, the likelihood of deformities, stillbirths, and early mortality in the offspring increases significantly. In related marriages, recessive genes often go into a homozygous state, as a result, certain anomalies develop. An example of this is the inheritance of hemophilia in the royal houses of Europe.

- hemophilic; - carrier woman

twin method

1 - monozygotic twins; 2 - dizygotic twins.

Children born at the same time are called twins. They are monozygotic(identical) and dizygotic(variegated).

Monozygotic twins develop from one zygote (1), which is divided into two (or more) parts during the crushing stage. Therefore, such twins are genetically identical and always of the same sex. Monozygotic twins are characterized by a high degree of similarity ( concordance) in many ways.

Dizygotic twins develop from two or more eggs that are simultaneously ovulated and fertilized by different spermatozoa (2). Therefore, they have different genotypes and can be either the same or different sex. Unlike monozygotic twins, dizygotic twins are characterized by discordance - dissimilarity in many ways. Data on the concordance of twins for some signs are given in the table.

signs	Concordance, %
	Monozygotic twins	dizygotic twins
Normal
Blood group (AB0)	100	46
eye color	99,5	28
Hair color	97	23
Pathological
Clubfoot	32	3
"Hare Lip"	33	5
Bronchial asthma	19	4,8
Measles	98	94
Tuberculosis	37	15
Epilepsy	67	3
Schizophrenia	70	13

As can be seen from the table, the degree of concordance of monozygotic twins for all the above characteristics is significantly higher than that of dizygotic twins, but it is not absolute. As a rule, the discordance of monozygotic twins occurs as a result of intrauterine development disorders of one of them or under the influence of the external environment, if it was different.

Thanks to the twin method, a person's hereditary predisposition to a number of diseases was clarified: schizophrenia, epilepsy, diabetes mellitus and others.

Observations on monozygotic twins provide material for elucidating the role of heredity and environment in the development of traits. Moreover, the external environment is understood not only as physical factors of the environment, but also as social conditions.

Cytogenetic method

Based on the study of human chromosomes in normal and pathological conditions. Normally, a human karyotype includes 46 chromosomes - 22 pairs of autosomes and two sex chromosomes. The use of this method made it possible to identify a group of diseases associated either with a change in the number of chromosomes or with changes in their structure. Such diseases are called chromosomal.

Blood lymphocytes are the most common material for karyotypic analysis. Blood is taken in adults from a vein, in newborns - from a finger, earlobe or heel. Lymphocytes are cultivated in a special nutrient medium, which, in particular, contains substances that “force” lymphocytes to intensively divide by mitosis. After some time, colchicine is added to the cell culture. Colchicine stops mitosis at the metaphase level. It is during metaphase that the chromosomes are most condensed. Next, the cells are transferred to glass slides, dried and stained with various dyes. Coloring can be a) routine (chromosomes stain evenly), b) differential (chromosomes acquire transverse striation, with each chromosome having an individual pattern). Routine staining allows you to identify genomic mutations, determine the group belonging of the chromosome, and find out in which group the number of chromosomes has changed. Differential staining allows you to identify chromosomal mutations, determine the chromosome to the number, find out the type of chromosomal mutation.

In cases where it is necessary to conduct a karyotypic analysis of the fetus, cells of the amniotic (amniotic) fluid are taken for cultivation - a mixture of fibroblast-like and epithelial cells.

Chromosomal diseases include: Klinefelter syndrome, Turner-Shereshevsky syndrome, Down syndrome, Patau syndrome, Edwards syndrome and others.

Patients with Klinefelter's syndrome (47, XXY) are always male. They are characterized by underdevelopment of the sex glands, degeneration of the seminiferous tubules, often mental retardation, high growth (due to disproportionately long legs).

Turner-Shereshevsky syndrome (45, X0) is observed in women. It manifests itself in slowing down puberty, underdevelopment of the gonads, amenorrhea (absence of menstruation), infertility. Women with Turner-Shereshevsky syndrome are small in stature, the body is disproportionate - the upper body is more developed, the shoulders are wide, the pelvis is narrow - the lower limbs are shortened, the neck is short with folds, the “Mongoloid” eye shape and a number of other signs.

Down syndrome is one of the most common chromosomal diseases. It develops as a result of trisomy on chromosome 21 (47; 21, 21, 21). The disease is easily diagnosed, as it has a number of characteristic features: shortened limbs, small skull, flat, wide nose bridge, narrow palpebral fissures with an oblique incision, the presence of a fold of the upper eyelid, mental retardation. Violations of the structure of internal organs are often observed.

Chromosomal diseases also occur as a result of changes in the chromosomes themselves. Yes, deletion R-arm of autosome number 5 leads to the development of the "cat's cry" syndrome. In children with this syndrome, the structure of the larynx is disturbed, and they early childhood have a peculiar "meowing" voice timbre. In addition, there is a retardation of psychomotor development and dementia.

Most often, chromosomal diseases are the result of mutations that have occurred in the germ cells of one of the parents.

Biochemical method

Allows you to detect metabolic disorders caused by changes in genes and, as a result, changes in the activity of various enzymes. Hereditary metabolic diseases are divided into diseases of carbohydrate metabolism (diabetes mellitus), metabolism of amino acids, lipids, minerals, etc.

Phenylketonuria refers to diseases of amino acid metabolism. The conversion of the essential amino acid phenylalanine to tyrosine is blocked, while phenylalanine is converted to phenylpyruvic acid, which is excreted in the urine. The disease leads to rapid development dementia in children. Early diagnosis and diet can stop the development of the disease.

Population-statistical method

It is a method of studying the distribution of hereditary traits (inherited diseases) in populations. An important point when using this method is statistical processing received data. Under population understand the totality of individuals of the same species, living in a certain territory for a long time, freely interbreeding with each other, having a common origin, a certain genetic structure and, to one degree or another, isolated from other such populations of individuals of a given species. A population is not only a form of existence of a species, but also a unit of evolution, since the basis of microevolutionary processes culminating in the formation of a species are genetic transformations in populations.

The study of the genetic structure of populations deals with a special section of genetics - population genetics. In humans, three types of populations are distinguished: 1) panmictic, 2) demes, 3) isolates, which differ from each other in number, frequency of intra-group marriages, the proportion of immigrants, and population growth. The population of a large city corresponds to the panmictic population. The genetic characteristics of any population includes the following indicators: 1) gene pool(the totality of genotypes of all individuals of a population), 2) gene frequencies, 3) genotype frequencies, 4) phenotype frequencies, marriage system, 5) factors that change gene frequencies.

To determine the frequencies of occurrence of certain genes and genotypes, hardy-weinberg law.

Hardy-Weinberg law

In an ideal population, from generation to generation, a strictly defined ratio of frequencies of dominant and recessive genes (1), as well as the ratio of frequencies of genotypic classes of individuals (2) is preserved.

p + q = 1, (1)
R 2 + 2pq + q 2 = 1, (2)

Where p— frequency of occurrence of the dominant gene A; q- the frequency of occurrence of the recessive gene a; R 2 - the frequency of occurrence of homozygotes for the dominant AA; 2 pq- frequency of occurrence of Aa heterozygotes; q 2 - the frequency of occurrence of homozygotes for the recessive aa.

The ideal population is a sufficiently large, panmictic (panmixia - free crossing) population, in which there is no mutation process, natural selection and other factors that disturb the balance of genes. It is clear that ideal populations do not exist in nature; in real populations, the Hardy-Weinberg law is used with amendments.

The Hardy-Weinberg law, in particular, is used to roughly count the carriers of recessive genes for hereditary diseases. For example, phenylketonuria is known to occur at a rate of 1:10,000 in a given population. Phenylketonuria is inherited in an autosomal recessive manner, therefore, patients with phenylketonuria have the aa genotype, that is q 2 = 0.0001. From here: q = 0,01; p= 1 - 0.01 = 0.99. Carriers of the recessive gene have the Aa genotype, that is, they are heterozygotes. The frequency of occurrence of heterozygotes (2 pq) is 2 0.99 0.01 ≈ 0.02. Conclusion: in this population, about 2% of the population are carriers of the phenylketonuria gene. At the same time, you can calculate the frequency of occurrence of homozygotes for the dominant (AA): p 2 = 0.992, just under 98%.

A change in the balance of genotypes and alleles in a panmictic population occurs under the influence of constantly acting factors, which include: the mutation process, population waves, isolation, natural selection, gene drift, emigration, immigration, inbreeding. It is thanks to these phenomena that an elementary evolutionary phenomenon arises - a change in the genetic composition of a population, which is the initial stage in the process of speciation.

Human genetics is one of the most intensively developing branches of science. She happens to be theoretical basis medicine, reveals the biological basis of hereditary diseases. Knowing the genetic nature of diseases allows you to make an accurate diagnosis in time and carry out the necessary treatment.

Go to lectures №21"Variability"

Hardy-Weinberg law

Population genetics deals with the genetic structure of populations.

The concept of "population" refers to a collection of freely interbreeding individuals of the same species, existing for a long time in a certain territory (part of the range) and relatively isolated from other populations of the same species.

The most important sign populations are relatively free interbreeding. If any isolation barriers arise that prevent free interbreeding, then new populations arise.

In humans, for example, in addition to territorial isolation, fairly isolated populations can arise on the basis of social, ethnic, or religious barriers. Since there is no free exchange of genes between populations, they can differ significantly in genetic characteristics. In order to describe the genetic properties of a population, the concept of a gene pool is introduced: the totality of genes found in a given population. In addition to the gene pool, the frequency of a gene or the frequency of an allele is also important.

Knowledge of how the laws of inheritance are implemented at the population level is fundamentally important for understanding the causes of individual variability. All patterns revealed in the course of psychogenetic studies refer to specific populations. In other populations, with a different gene pool and different gene frequencies, different results may be obtained.

The Hardy-Weinberg law is the basis of mathematical constructions of population genetics and modern evolutionary theory. Formulated independently by the mathematician G. Hardy (England) and physician W. Weinberg (Germany) in 1908. This law states that the frequencies of alleles and genotypes in a given population will remain constant from generation to generation under the following conditions:

1) the number of individuals in the population is large enough (ideally, infinitely large),

2) mating occurs randomly (i.e., panmixia occurs),

3) there is no mutation process,

4) there is no exchange of genes with other populations,

5) there is no natural selection, i.e. individuals with different genotypes are equally fertile and viable.

Sometimes this law is formulated differently: in an ideal population, the frequencies of alleles and genotypes are constant. (Because the above conditions for the fulfillment of this law are the properties of an ideal population.)

The mathematical model of the law corresponds to the formula:

It is derived based on the following reasoning. As an example, let's take the simplest case - the distribution of two alleles of one gene. Let two organisms be the founders of a new population. One of them is homozygous dominant (AA) and the other homozygous recessive (aa). Naturally, all their offspring in F 1 will be uniform and will have the genotype (Aa). Further individuals F 1 will interbreed with each other. Let us denote the frequency of occurrence of the dominant allele (A) by the letter p, and the recessive allele (a) by the letter q. Since the gene is represented by only two alleles, the sum of their frequencies is equal to one, i.e. p + q = 1. Consider all the eggs in this population. The proportion of eggs carrying the dominant allele (A) will correspond to the frequency of this allele in the population and, therefore, will be p. The proportion of eggs carrying the recessive allele (a) will correspond to its frequency and will be q. Having carried out similar reasoning for all spermatozoa of the population, we will come to the conclusion that the proportion of spermatozoa carrying the allele (A) will be p, and those carrying the recessive allele (a) will be q. Now we will compose the Punnett lattice, while writing the types of gametes we will take into account not only the genomes of these gametes, but also the frequencies of the alleles they carry. At the intersection of the rows and columns of the lattice, we will get the genotypes of the offspring with coefficients corresponding to the frequencies of occurrence of these genotypes.

It can be seen from the above lattice that in F 2 the frequency of dominant homozygotes (AA) is p, the frequency of heterozygotes (Aa) is 2pq, and the frequency of recessive homozygotes (aa) is q. Since the given genotypes represent all possible options genotypes for the case we are considering, then the sum of their frequencies should be equal to one, i.e.

The main application of the Hardy-Weinberg law in the genetics of natural populations is the calculation of allele and genotype frequencies.

Let's consider an example of using this law in genetic calculations. It is known that one person out of 10 thousand is an albino, while the sign of albinism in humans is determined by one recessive gene. Let's calculate what is the proportion of hidden carriers of this trait in the human population. If one person out of 10 thousand is an albino, then this means that the frequency of recessive homozygotes is 0.0001, i.e. q 2 \u003d 0.0001. Knowing this, it is possible to determine the frequency of the allele of albinism q, the frequency of the dominant allele of normal pigmentation p and the frequency of the heterozygous genotype (2pq). People with such a genotype will just be hidden carriers of albinism, despite the fact that phenotypically this gene will not appear in them and they will have normal skin pigmentation.

From the above simple calculations, it can be seen that, although the number of albinos is extremely small - only one person per 10 thousand, the albinism gene carries a significant number of people - about 2%. In other words, even if a trait is phenotypically manifested very rarely, then there is a significant number of carriers of this trait in the population, i.e. individuals with this gene in the heterozygote.

Thanks to the discovery of the Hardy-Weinberg law, the process of microevolution became available for direct study: its course can be judged by changes from generation to generation of gene frequencies (or genotypes). Thus, despite the fact that this law is valid for an ideal population, which does not and cannot exist in nature, it is of great practical importance, since it makes it possible to calculate gene frequencies that change under the influence of various factors of microevolution.

EXAMPLES OF SOLVING PROBLEMS

1. Albinism in rye is inherited as an autosomal recessive trait. In a plot of 84,000 plants, 210 turned out to be albinos. Determine the frequency of the albinism gene in rye.

Solution

Due to the fact that albinism in rye is inherited as an autosomal recessive trait, all albino plants will be homozygous for the recessive gene - aa. Their frequency in the population (q 2 ) equals 210/84000 = 0.0025. Recessive gene frequency A will be equal to 0.0025. Hence, q = 0,05.

Answer:0,05

2. In cattle, the red color does not completely dominate over the white (hybrids have a roan color). 4169 red, 756 white and 3708 roan animals were found in the area. What is the frequency of livestock color genes in this area?

Solution.

If the gene for the red suit of animals is denoted by A,
and the white gene A, then in red animals the genotype will be AA(4169), in roans Ah(3780), for whites - aa(756). A total of 8705 animals were registered. You can calculate the frequency of homozygous red and white animals in fractions of a unit. The frequency of white animals will be 756: 8705 = 0.09. Hence q 2 =0.09 . Recessive gene frequency q= = 0.3. gene frequency A will p = 1 - q. Therefore, R= 1 - 0,3 = 0,7.

Answer:R= 0.7, gene q = 0,3.

3. In humans, albinism is an autosomal recessive trait. The disease occurs with a frequency of 1/20,000. Determine the frequency of heterozygous carriers of the disease in the area.

Solution.

Albinism is inherited recessively. Value 1/20000 -
This q 2 . Therefore, the frequency of the gene A will: q = 1/20000 =
= 1/141. The p gene frequency will be: R= 1 - q; R= 1 - 1/141 = 140/141.

The number of heterozygotes in the population is 2 pq= 2 x (140/141) x (1/141) = 1/70. Because in a population of 20,000 people, then the number of heterozygotes in it is 1/70 x 20,000 = 286 people.

Answer: 286 people

4. Congenital dislocation of the hip in humans is inherited as a sutosomal dominant trait with a penetrance of 25%. The disease occurs with a frequency of 6:10,000. Determine the number of heterozygous carriers of the gene for congenital hip dislocation in the population.

Solution.

The genotypes of individuals with congenital hip dislocation, AA And Ah(dominant inheritance). Healthy individuals have the aa genotype. From the formula R 2 + 2pq+. q 2 =1 it is clear that the number of individuals carrying the dominant gene is (p 2 + 2pq). However, the number of patients given in the problem of 6/10000 represents only one fourth (25%) of gene A carriers in the population. Hence, R 2 + 2pq =(4 x 6)/10,000 = 24/10,000. Then q 2 (the number of individuals homozygous for the recessive gene) is 1 - (24/10000) = 9976/10000 or 9976 people.

Answer: 9976 people

4. Allele frequencies p = 0.8 and g = 0.2 are known in the population. Determine genotype frequencies.

Given:		Solution:
p=0.8 g = 0.2 p2-? g2 - ? 2pg - ?		p2 = 0.64 g 2 \u003d 0.04 2pg = 0.32

Answer: genotype frequency AA– 0.64; genotype aa– 0.04; genotype Ah – 0,32.

5. The population has the following composition: 0.2AA, 0,3 Ahand 0.50aa. Find Allele FrequenciesAAndA.

Given:		Solution:
p 2 \u003d 0.2 g 2 \u003d 0.3 2pg = 0.50 p-? g-?		p=0.45 g = 0.55

Answer: allele frequency A– 0.45; allele A – 0,55.

6. In a herd of cattle, 49% of animals are red (recessive) and 51% black (dominant). What percentage of homo- and heterozygous animals are in this herd?

Given:		Solution:
g 2 \u003d 0.49 p 2 + 2pg = 0.51 p-? 2pg - ?		g = 0.7 p = 1 – g = 0.3 p2 = 0.09 2pg = 0.42

Answer: heterozygotes 42%; homozygous recessive - 49%; homozygotes for the dominant - 9%.

7. Calculate genotype frequenciesAA, AhAndaa(in%) if individualsaamake up 1% of the population.

Given:		Solution:
g 2 = 0,01 p2-? 2pg - ?		g = 0.1 p = 1 - g = 0.9 2pg = 0.18 p2 = 0.81

Answer: in the population 81% of individuals with the genotype AA, 18% with genotp Ah and 1% with the genotype aa.

8. When examining the population of Karakul sheep, 729 long-eared individuals (AA), 111 short-eared (Aa) and 4 earless (aa) individuals were identified. Calculate observed phenotype frequencies, allele frequencies, 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 could be determined based on the available data.

To do this, you just need to find the sum of all individuals in the population (it is equal to 844), find the proportion of long-eared, short-eared and earless first in percent (86.37, 13.15 and 0.47, respectively) and in frequency fractions (0.8637, 0.1315 and 0.00474).

But the task says to apply the Hardy-Weinberg formula for calculating genotypes and phenotypes and, moreover, to calculate the allele frequencies of genes A and a. So, to calculate the frequencies of alleles of genes, one cannot do without the Hardy-Weinberg formula.

Let us 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. The sum of allelic gene frequencies p + q = 1.

Since, according to the Hardy-Weinberg formula p 2 AA + 2pqAa + q 2 aa = 1, we have that the frequency of occurrence of the earless q 2 is 0.00474, then extracting 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 of the dominant allele A. It is equal to 1 - 0.06884 = 0.93116.

Now, using the formula, we can again calculate the frequencies of occurrence of long-eared (AA), earless (aa) and short-eared (Aa) individuals. Long-eared with the AA genotype will be p 2 = 0.931162 = 0.86706, earless with the aa genotype will be q 2 = 0.00474 and short-eared with the Aa genotype will be 2pq = 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) .

TASKS FOR INDEPENDENT SOLUTION

1. One of the forms of glucosuria is inherited as an autosomal recessive trait and occurs with a frequency of 7:1000000. Determine the frequency of occurrence of heterozygotes in the population.

2. General albinism (milky white skin color, 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.

3. In rabbits, chinchilla hair color (Cch gene) dominates over albinism (Ca gene). CchCa heterozygotes are light gray in color. On a rabbit farm, albinos appeared among the young chinchilla rabbits. Of the 5400 rabbits, 17 turned out to be albinos. Using the Hardy-Weinberg formula, determine how many homozygous chinchilla-colored rabbits were obtained.

4. The population of Europeans according to the Rh blood group system contains 85% of Rh-positive individuals. Determine the saturation of the population with a recessive allele.

5. 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 for the analyzed trait, based on these data.

Solution 1 Let us designate 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 will be denoted by A.

Healthy individuals in the human population have the AA and Aa genotypes; diseased individuals have the genotype only aa.

Let us denote the frequency of occurrence of the recessive allele a by the letter q, and the frequency of 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 = 0.00264575

Since p + q = 1, then p = 1 - q = 0.9973543, and p2 = 0.9947155

Now substituting the values ​​of p and q into the formula: p2AA + 2pqAa + q2aa = 1,
let's find the frequency of occurrence of heterozygous individuals 2pq in the human population: 2pq \u003d 1 - p 2 - q 2 \u003d 1 - 0.9947155 - 0.000007 \u003d 0.0052775.

Solution 2 Since this trait is recessive, diseased organisms will have the aa genotype - their frequency is 1: 20,000 or 0.00005.
The allele frequency a will be the square root of this number, that is, 0.0071. The allele frequency A will be 1 - 0.0071 = 0.9929, and the frequency of healthy AA homozygotes will be 0.9859. The frequency of all heterozygotes 2Aa = 1 - (AA + aa) = 0.014 or 1.4% .

Solution 3 Let's take 5400 pieces of all rabbits as 100%, then 5383 rabbits (the sum of AA and Aa genotypes) will be 99.685% or in parts it will be 0.99685.

q 2 + 2q (1 - q) \u003d 0.99685 - this 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 = 1 , we find (1 - q) 2 = 1 - 0.99685 = 0.00315 - this is the frequency of occurrence of albino rabbits with aa genotype. We find what the value 1 - q is equal to. This is the square root of 0.00315 = 0.056. And q then equals 0.944.

q 2 is equal to 0.891, and this is the proportion of homozygous chinchillas with the AA genotype. Since this value in% will be 89.1% of 5400 individuals, the number of homozygous chinchillas will be 4811 pcs. .

Solution 4 We know that the allelic gene responsible for the manifestation of Rh positive blood is dominant R (we denote the frequency of its occurrence by the letter p), and Rh negative is recessive r (we denote its frequency by the letter q).

Since the task says that p 2 RR + 2pqRr accounts for 85% of people, then the share of Rh-negative q 2 rr phenotypes will account for 15% or their frequency of occurrence will be 0.15 of all people of the European population.

Then the frequency of occurrence of the allele r or “saturation of the population with a recessive allele” (indicated by the letter q) will be the square root of 0.15 = 0.39 or 39%.

Solution 5 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 for the analyzed trait, 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%, 5 times more men are actually carriers of gout genes, 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 the AA + 2Aa genotypes in the population, which means that 95 out of 100 will be with the aa genotype.

If the frequency of occurrence of organisms with genotypes aa 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 = 0.975. Then the frequency of occurrence of the dominant allele "A" in this population is 1 - 0.975 = 0.005 .

SOLUTION OF TYPICAL TASKS

Task 1. In the South American jungle lives a population of 127 people (including children). The frequency of blood type M is 64% here. Is it possible to calculate the N and MN blood group frequencies in this population?

Solution. For a small population, the mathematical expression of the Hardy-Weinberg law cannot be applied, so it is impossible to calculate the frequencies of gene occurrence.

Task 2. Tay-Sachs disease, caused by an autosomal recessive gene, is incurable; people suffering from this disease die in childhood. In one of the large populations, the birth rate of sick children is 1:5000. Will the concentration of the pathological gene and the frequency of this disease change in the next generation of this population?

Solution

We produce a mathematical record of the Hardy-Weinberg law

p + q - 1, p 2 .+ 2pq + q 2 = 1.

p is the frequency of occurrence of gene A;

q is the frequency of occurrence of gene a;

p 2 - frequency of occurrence of dominant homozygotes

2pq - frequency of occurrence of heterozygotes (Aa);

q 2 - the frequency of occurrence of recessive homozygotes (aa).

From the condition of the problem, according to the Hardy-Weinberg formula, we know the frequency of occurrence of sick children (aa), i.e. q 2 = 1/5000.

The gene that causes this disease will pass to the next generation only from heterozygous parents, so it is necessary to find the frequency of occurrence of heterozygotes (Aa), i.e. 2pq.

q \u003d 1/71, p \u003d l-q - 70/71, 2pq \u003d 0.028.

We determine the concentration of the gene in the next generation. It will be in 50% of gametes in heterozygotes, its concentration in the gene pool is about 0.014. The probability of the birth of sick children q 2 = 0.000196, or 0.98 per 5000 population. Thus, the concentration of the pathological gene and the frequency of this disease in the next generation of this population will practically not change (the decrease is insignificant).

Task 3. Congenital hip dislocation is dominantly inherited, the average penetrance of the gene is 25%. The disease occurs with a frequency of 6:10,000 (V. P. Efroimson, 1968). Determine the number of homozygous individuals for the recessive gene.

Solution. We arrange the condition of the problem in the form of a table:

hip dislocation

Thus, from the conditions of the problem, according to the Hardy-Weinberg formula, we know the frequency of occurrence of the AA and Aa genotypes, i.e. p 2 + 2pq. It is necessary to find the frequency of occurrence of the aa genotype, i.e. q 2 .

From the formula p 2 -t- 2pq + q 2 \u003d l, it is clear that the number of individuals homozygous for the recessive gene (aa) q 2 \u003d 1 - (p 2 + 2pq). However, the number of patients given in the problem (6:10,000) is not p 2 + 2pq, but only 25% of gene A carriers, while the true number of people with this gene is four times greater, i.e. 24: 10,000 Therefore, p 2 + 2pq = 24:10 000. Then q 2 (number

individuals homozygous for the recessive gene) is 9976:10,000.

Task 4. The Kidd blood group system is determined by the allelic genes Ik a and Ik b. The Ik a gene is dominant over the Ik b gene and individuals who have it are kidd-positive. The frequency of the Ik a gene among the population of Krakow is 0.458 (V. Socha, 1970).

The frequency of kidd-positive people among blacks is 80%. (K. Stern, 1965). Determine the genetic structure of the population of the city of Krakow and blacks according to the Kidd system.

Solution. We arrange the condition of the problem in the form of a table:

We make a mathematical record of the Hardy-Weinberg law: - p + q \u003d I, p 2 + 2pq + q 2 \u003d 1.

p is the frequency of occurrence of the gene Ik α ;

q - frequency of occurrence of the gene Ik β ; . p 2 - frequency of occurrence of dominant homozygotes (Ik α lk α);

2pq - frequency of occurrence of heterozygotes (Ik α Ik β);

q 2 - the frequency of occurrence of recessive homozygotes (Ik β Ik β).

Thus, from the condition of the problem, according to the Hardy-Weinberg formula, we know the frequency of occurrence of the dominant gene in the Krakow population - p = 0.458 (45.8%). We find the frequency of occurrence of the recessive gene: q = 1 - 0.458 = 0.542 (54.2%). We calculate the genetic structure of the Krakow population: the frequency of occurrence of dominant homozygotes - p 2 = 0.2098 (20.98%); frequency of occurrence of heterozygotes - 2pq = 0.4965 (49.65%); the frequency of occurrence of recessive homozygotes - Q 2 = 0.2937 (29.37%).

For blacks, from the condition of the problem, we know the frequency of occurrence of dominant homozygotes and heterozygotes (with

dominant sign), i.e. p 2 +2pq=0.8. According to the Hardy-Weinberg formula, we find the frequency of occurrence of recessive homozygotes (Ik β Ik β): q 2 =1-p 2 +2pq=0.2 (20%). Now we determine the frequency of the recessive gene Ik β: q=0.45 (45%). We find the frequency of occurrence of the gene Ik α: p=1-0.45=0.55 (55%); the frequency of occurrence of dominant homozygotes (Ik α Ik α): p 2 = 0.3 (30%); frequency of occurrence of heterozygotes (Ik α Ik β): 2pq = 0.495 (49.5%).

TASKS FOR SELF-CONTROL

Task 1. Children with phenylketonuria are born with a frequency of 1:10,000 newborns. Determine the percentage of heterozygous gene carriers.

Task 2. General albinism (milky white skin color, 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.

Task 3. Hereditary methemoglobinemia, an autosomal recessive trait, occurs in the Eskimos of Alaska with a frequency of 0.09%. Determine the genetic structure of the population for this trait.

Task 4. People with blood type N among the population of Ukraine make up 16%. Determine the frequency of groups M and MN.

Task 5. In Papuans, the frequency of blood type N is 81%. Determine the frequency of M and MN groups in this population.

Task 6. When examining the population of southern Poland, individuals with blood types were found: M - 11163, MN - 15267, N - 5134. Determine the frequency of genes L N and L M among the population of southern Poland.

Task 7. The incidence of gout is 2%; it is due to a dominant autosomal gene. According to some data (V.P. Efroimson, 1968), the gout gene penetrance in men is 20%, and in women - 0%.

Determine the genetic structure of the population according to the analyzed trait.

Task 8. In the US, about 30% of the population perceives the bitter taste of phenylthiocarbamide (PTC), while 70% do not. The ability to taste FTK is determined by the recessive gene a. Determine the frequency of alleles A and a in this population.

Task 9. One of the forms of fructosuria is inherited as an autosomal recessive trait and occurs with a frequency of 7: 1,000,000 (V. P. Efroimson, 1968). Determine the frequency of occurrence of heterozygotes in the population.

Task 10. Determine the frequency of occurrence of albinos in a large African population, where the concentration of the pathological recessive gene is 10%.

Task 11. Aniridia (absence of the iris) is inherited as an autosomal dominant trait and occurs with a frequency of 1:10,000 (V.P. Efroimson, 1968). Determine the frequency of occurrence of heterozygotes in the population.

Task 12. Essential pentosuria (urinary excretion of L-xylulose) is inherited as an autosomal recessive trait and occurs with a frequency of 1: 50,000 (L.O. Badalyan, 1971). Determine the frequency of occurrence of dominant homozygotes in the population.

Task 13. Alkaptonuria (urinary excretion of homogentisic acid, staining of cartilaginous tissues, development of arthritis) is inherited as an autosomal recessive trait with a frequency of 1:100,000 (V.P. Efroimson, 1968). Determine the frequency of occurrence of heterozygotes in the population.

Task 14. Blood groups according to the system of antigens M and N (M, MN, N) are determined by the codominant genes L N and L M . The frequency of occurrence of the L M gene in the white population of the United States is 54%, in Indians - 78%, in Greenland Eskimos - 91%, in Australian Aborigines - 18%. Determine the frequency of occurrence of the MN blood group in each of these populations.

Task 15. One grain of wheat, heterozygous for gene A, accidentally fell on a deserted island. The grain sprouted and gave rise to a series of generations that reproduce by self-pollination. What will be the proportion of heterozygous plants among the representatives of the first, second, third; fourth generation, if the trait determined by the genome does not affect the survival of plants and their reproduction?

Task 16. Albinism in rye is inherited as an autosomal recessive trait. In the surveyed area, 210 albinos were found among 84,000 plants. Determine the frequency of occurrence of the albinism gene in rye.

Problem 17. On one of the islands, 10,000 foxes were shot. 9991 of them were red (dominant) and 9 white (recessive). Determine the frequency of occurrence of genotypes of homozygous red foxes, heterozygous red foxes and whites in this population.

Task 18. In a large population, the frequency of the gene for color blindness (a recessive, X-linked trait) among men is 0.08. Determine the frequency of occurrence of genotypes of dominant homozygotes, heterozygotes, recessive homozygotes in women of this population.

Problem 19. In shothorn cattle, color is inherited as an autosomal trait with incomplete dominance: hybrids from crossing red and white animals have a roan color. In area N, specialized in the breeding of shorthorns, 4169 red animals, 3780 roans and 756 whites were registered. Determine the frequency of the genes that cause the red and white color of cattle in the area.

HUMAN GENETICS

SOLUTION OF TYPICAL TASKS

Task 1. Determine the type of inheritance

Solution. The trait occurs in every generation. This immediately eliminates the recessive type of inheritance. Since this trait occurs in both men and women, this excludes the hollandic type of inheritance. There are two possible types of inheritance: autosomal dominant and sex-linked dominant, which are very similar. A man II - 3 has daughters both with this trait (III-1, III-5, III-7) and without it (III-3), which excludes a sex-linked dominant type of inheritance. So, in this pedigree - an autosomal dominant type of inheritance.

Task 2

Solution. The trait does not occur in every generation. This excludes the dominant type of inheritance. Since the trait occurs in both males and females, this rules out a hollandic type of inheritance. To exclude a sex-linked recessive type of inheritance, it is necessary to consider the marriage pattern of III-3 and III-4 (the trait does not occur in a man and a woman). If we assume that the male genotype is X A Y, and the female genotype is X A X a, they cannot have a daughter with this trait (X a X a), and in this pedigree there is a daughter with this trait - IV-2. Given the occurrence of the trait equally in both men and women and the case of closely related marriage, it can be concluded that this pedigree has an autosomal recessive type of inheritance.

Task 3. The concordance of monozygotic twins by body weight is 80%, and that of dizygotic twins is 30%. What is the ratio of hereditary and environmental factors in the formation of a trait?

Solution. Using the Holzinger formula, we calculate the heritability coefficient:

KMB%-KDB%

80% - 30%

Since the heritability coefficient is 0.71, the genotype plays an important role in the formation of the trait.

TASKS FOR SELF-CONTROL

Task 1. Determine the type of inheritance.

Task 2. Determine the type of inheritance.

Task 3. Determine the type of inheritance.

Task 4. ABO blood groups in monozygotes

twins coincide in 100% of cases, and in dizygotic twins - in 40%. What determines the coefficient of heritability - environment or heredity?

Task 5. Rickets resistant to vitamin D (hypophosphatemia) is a hereditary disease caused by a dominant gene located on the X chromosome. In a family where the father suffers from this disease and the mother is healthy, there are 3 daughters and 3 sons. How many of them can be sick?

Task 6. Is the protein composition of two monozygotic twins the same if there were no mutations in their cells?

Task 7. Which of the following features characterize the autosomal dominant type of inheritance: a) the disease is equally common in women and men; b) the disease is transmitted from parents to children in each generation; c) all daughters of a sick father are sick; d) the son never inherits the disease from his father; e) Are the parents of the sick child healthy?

Task 8. Which of the following features characterize the autosomal recessive type of inheritance: a) the disease is equally common in women and men; b) the disease is transmitted from parents to children in each generation; c) all daughters of a sick father are sick; d) parents are blood relatives; e) Are the parents of the sick child healthy?

Task 9. Which of the following features characterize the dominant, X-linked type of inheritance: a) the disease is equally common in women and men; b) the disease is transmitted from parents to children in each generation; c) all daughters of a sick father are sick; d) the son never inherits the disease from his father; e) if the mother is sick, then regardless of gender, the probability of having a sick child is 50%?

Within the gene pool of a population, the proportion of genotypes containing different alleles of one gene; subject to certain conditions from generation to generation does not change. 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 infinite a large number 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.

The Hardy-Weinberg equation in solving genetic problems

It is well known that this law is applicable only for ideal populations: a sufficiently high number of individuals in the population; the population should be panmix when there is no restriction on the free choice of a sexual partner; there should be practically no mutation of the trait under study; there is no inflow and outflow of genes and there is no natural selection.

The Hardy-Weinberg law is formulated as follows:

in an ideal population, the ratio of allele frequencies of genes and genotypes from generation to generation is a constant value and corresponds to the equation:

p 2 +2pq + q 2 = 1

Where p 2 is the proportion of homozygotes for one of the alleles; p is the frequency of this allele; q 2 - proportion of homozygotes for the alternative allele; q is the frequency of the corresponding allele; 2pq is the proportion of heterozygotes.

What does “ratio of allele frequencies of genes” and “ratio of genotypes” mean - constant values? What are these values?

Let the frequency of occurrence of any gene in the dominant state (A) be equal to p, and the recessive allele (a) of the same gene is equal to q(it is possible and vice versa, or it is possible in general with one letter, expressing one designation from another) and realizing that the sum of the frequencies of the dominant and recessive alleles of one gene in the population is 1, we get the first equation:

1) p + q = 1

Where does the Hardy-Weinberg equation come from? You remember that when monohybrid crossing of heterozygous organisms with genotypes Aa x Aa according to Mendel's second law in the offspring, we will observe the appearance of different genotypes in the ratio 1AA: 2 Aa: 1aa.

Since the frequency of occurrence of the dominant allelic gene A is denoted by the letter p, and the recessive allele a by the letter q, then the sum of the frequencies of occurrence of the genotypes of organisms themselves (AA, 2Aa and aa) that have the same allelic genes A and a will also be equal to 1, then :

2) p 2 AA + 2pqAa + q 2 aa =1

In tasks in population genetics, as a rule, it is required:
a) find the frequencies of occurrence of each of the allelic genes according to the known ratio of the frequencies of the genotypes of individuals;

B) or vice versa, find the frequency of occurrence of any of the genotypes of individuals according to the known frequency of occurrence of the dominant or recessive allele of the trait under study.

So, substituting the known value of the frequency of occurrence of one of the alleles of the gene into the first formula and finding the value of the frequency of occurrence of the second allele, we can always use the Hardy-Weinberg equation to find the frequencies of occurrence of the various genotypes of the offspring themselves.

Usually some actions (because of their obviousness) are solved in the mind. But in order to make clear what is already obvious, one must have a good understanding of what the letter designations in the Hardy-Weinberg formula are.

The provisions of the Hardy-Weinberg law also 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 = 1;

P 2 AA + q 2 a1a1 + r 2 a2a2 + 2pqAa1 + 2prAa2 + 2qra1a2 \u003d 1.

"In a population of an infinite number of freely interbreeding individuals V no mutations, selective migration organisms with different genotypes and pressure of natural selection original allele frequencies persist from generation to generation.”

Suppose that in the gene pool of a population that satisfies the described conditions, a certain gene is represented by the alleles A 1 and A 2, found with a frequency of p and q. Since there are no other alleles in this gene pool, then p + q \u003d 1. In this case, q \u003d 1-p.

Accordingly, individuals of this population form p gametes with the A 1 allele and q gametes with the A 2 allele. If crossings occur randomly, then the proportion of germ cells that combine with gametes A 1 is equal to p, and the proportion of germ cells that combine with gametes A 2 is q. The generation F 1 arising as a result of the described reproduction cycle is formed by the genotypes A l A 1, A 1 A 2, A 2 A 2, the number of which correlates as (p + q) (p + q) = p 2 + 2pq + q 2 (Fig. .10.2). Upon reaching puberty, individuals of AlAi and ArA2 each 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 equal frequency 2pq /2.

Rice. Regular distribution of genotypes in a number of generations depending on the frequency of formation of gametes of different types (Hardy-Weinberg law)

Thus, the proportion of gametes A 1 in generation F 1 will be p 2 + 2pq / 2 = p 2 + p (1 - p) \u003d p, and the proportion of gametes A 2 will be equal to q 2 + 2pq / 2 \u003d q 2 + + q (l -q ) = q .

Since the frequencies of gametes with different alleles in the generation fi are not changed in comparison with the parental 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 made for loci with any number of alleles. The conservation of allele frequencies is based on statistical patterns random events in large samples.

The Hardy-Weinberg equation, as discussed above, is valid for autosomal genes. For sex-linked genes, 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: р 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 are reproduced with a frequency equal to the frequency of the corresponding alleles in females in the previous generation: p and q. It follows from this that phenotypes determined by recessive alleles of X-linked genes are more common in males than in females.

So, with an allele frequency of hemophilia equal to 0.0001, this disease in men of this population is observed 10,000 times more often than in women (1 in 10 thousand in the former and 1 in 100 million in the latter).

Another consequence of the general order is that in the case of an inequality in the allele frequency in males and females, the difference between the frequencies in the next generation is halved, and the sign of this difference changes. It usually takes several generations for an equilibrium state of frequencies to occur in both sexes. The specified state for autosomal genes is achieved in one generation.

The Hardy-Weinberg law describes the conditions population genetic stability. A population whose gene pool does not change over generations is called Mendelian. The genetic stability of Mendelian populations puts them outside the evolutionary process, since under such conditions the action of natural selection is suspended. The identification of Mendelian populations is of purely theoretical significance. These populations do not occur in nature. The Hardy-Weinberg law lists the conditions that naturally change the gene pools of populations. For example, factors that limit free crossing (panmixia), such as the finite number of organisms in a population, isolation barriers that prevent random selection of marriage pairs, lead to this result. Genetic inertia is also overcome through mutations, the influx into or out of the population of individuals with certain genotypes, and selection.

Examples of solutions to some tasks using the Hardy-Weinberg equation.

Task 1. In the human population, the number of individuals with brown eyes is 51%, and with blue - 49%. Determine the percentage of dominant homozygotes in this population.

The complexity of solving such tasks is in their apparent simplicity. Since there is so little data, then the solution should seem to be very short. It turns out not very much.

According to the condition of such tasks, we are usually given information about the total number of phenotypes of individuals in the population. Since the phenotypes of individuals in a population with dominant traits can be represented by both homozygous AA individuals and heterozygous Aa individuals, in order to determine the frequencies of occurrence of some specific genotypes of individuals in this population, it is necessary to first calculate the frequencies 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 brown-eyed trait A, and the allelic gene responsible for the manifestation of blue eyes, respectively, a. Then brown-eyed people in the studied population will be people with the AA genotype (dominant homozygotes, the proportion of which must be found according to the condition of the problem), and - Aa heterozygotes), and blue-eyed - only aa (recessive homozygotes).

By the condition of the problem, we know that the number of people with the AA and Aa genotypes is 51%, and the number of people with the aa genotype is 49%. How, based on these statistics (a large sample should be representative), can one calculate the percentage of brown-eyed people with only the AA genotype?

To do this, we calculate the frequencies of occurrence of each of the allelic genes A and a in a given population of people. The Hardy-Weinberg law, applied to large freely interbreeding populations, will just allow us to do this.

Denoting the frequency of occurrence of allele A in a given population by the letter q, we have the frequency of occurrence of the allelic gene a = 1 - q. (It would be possible to designate the frequency of occurrence of the allelic gene a with a separate letter, as in the text above - this is more convenient for anyone). Then the Hardy-Weinberg formula itself for calculating the frequencies of genotypes in monohybrid crossing 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 = 1.

Well, now everything is simple, you probably all guessed what we know in this equation, and what should be found?

(1 - q) 2 = 0.49 is the frequency of people with blue eyes.

Find the value of q: 1 - q = square root of 0.49 = 0.7; q = 1 - 0.7 = 0.3, then q2 = 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 equal to 9%.

Problem 2. In red clover, late maturity dominates over early maturity and is inherited monogenously. During approbation, it was found that 4% of plants belong to the early-ripening type of clover, what part of the late-ripening plants are heterozygotes?

In this context, approbation means assessing the purity of a variety. But what if the variety is not a pure line like Mendel's pea varieties, for example. Theoretically, “yes”, but in practice (the fields are large - these are not experimental plots of the brilliant Mendel) in each production variety there may be some amount of “garbage” alleles of genes.

IN this case with a late maturing clover variety, if the variety were pure, only plants with the AA genotype would be present. But the variety turned out to be not very pure at the time of testing (testing), since 4% of the individuals were early ripe plants with the aa genotype. This means that alleles a have been “crawled” into this variety.

So, since they are “messed up”, then individuals should also be present in this variety, although they are late-ripening in terms of phenotype, but heterozygous with the Aa genotype - do we need to determine their number?

According to the condition of the problem, 4% of individuals with the aa genotype will make up 0.04 of the entire variety. In fact, this is q 2, which means that the frequency of occurrence of the recessive allele a is equal to q \u003d 0.2. Then the frequency of occurrence of the dominant allele A is p = 1 - 0.2 = 0.8.

Hence the number of late-ripening homozygotes p2 = 0.64 or 64%. Then the number of Aa heterozygotes will be 100% - 4% - 64% = 32%. Since there are 96% of late-ripening plants in total, the proportion of heterozygotes among them will be: 32 x 100: 96 = 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 earless (aa) individuals were identified. Calculate observed phenotype frequencies, allele frequencies, 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 could be determined based on the available data. To do this, you just need to find the sum of all individuals in the population (it is equal to 844), find the proportion of long-eared, short-eared and earless first in percent (86.37, 13.15 and 0.47, respectively) and in frequency fractions (0.8637, 0.1315 and 0.00474).

But the task says to apply the Hardy-Weinberg formula for calculating genotypes and phenotypes and, moreover, to calculate the allele frequencies of genes A and a. So, to calculate the frequencies of alleles of genes, one cannot do without the Hardy-Weinberg formula.

Please note that in this task, unlike the previous one, to denote the frequencies of allelic genes, we will use the notation not as in the first task, but as discussed above in the text. It is clear that the result will not change from this, but in the future you will be free to use any of these notation methods that seems more convenient for you to understand and carry out the calculations themselves.

Let us 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 allelic gene frequencies is p + q = 1.

Since, according to the Hardy-Weinberg formula p 2 AA + 2pqAa + q 2 aa = 1, we have that the frequency of occurrence of the earless q2 is 0.00474, then by extracting the square root of 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 of the dominant allele A. It is equal to 1 - 0.06884 = 0.93116.

Now, using the formula, we can again calculate the frequencies of occurrence of long-eared (AA), earless (aa) and short-eared (Aa) individuals. Long-eared with the AA genotype will be p 2 = 0.931162 = 0.86706, earless with the aa genotype will be q 2 = 0.00474 and short-eared with the Aa genotype will be 2pq = 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 found to be albinos because their recessive genes are in the homozygous state. Determine the frequencies of alleles A and a, as well as the frequency of heterozygous plants.

Let us 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 = 1 give us to apply it to this problem?

Since the total number of all individuals of this rye population is known to us 84,000 plants, and in parts this is 1, then the proportion of homozygous albino individuals with the aa genotype equal to q2, of which there are only 210 pieces, will be q2 = 210: 84000 = 0.0025, then q = 0.05; p = 1 - q = 0.95 and then 2pq = 0.095.

Answer: allele frequency a - 0.05; allele frequency A - 0.95; the frequency of heterozygous plants with the Aa genotype will be 0.095.

Task 5. They raised chinchilla rabbits, but got a marriage in the form of albinos

In rabbits, chinchilla hair color (Cch gene) dominates over albinism (Ca gene). CchCa heterozygotes are light gray in color. On a rabbit farm, albinos appeared among the young chinchilla rabbits. Of the 5400 rabbits, 17 turned out to be albinos. Using the Hardy-Weinberg formula, determine how many homozygous chinchilla-colored rabbits were obtained.

Do you think that the resulting 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 apply the Hardy-Weinberg formula in the calculations. In order to use it correctly, we must clearly understand what is given to us and what we need to find.

Only for the convenience of registration, we will denote the genotype of chinchillas AA (we will have to determine the number of them), the albinos genotype aa, then the genotype of heterozygous seryachki will be denoted Aa.

If you “add up” all 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 can we now, not knowing how many heterozygous gray rabbits with the Aa genotype, 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 one, only there we were given the results of calculations in the population of people of brown-eyed and blue-eyed individuals in%, and here, in fact, we know the number of albinos rabbits 17 pieces and all homozygous chinchillas and heterozygous grays in total : 5400 - 17 = 5383 pieces.

Let's take 5400 pieces of all rabbits as 100%, then 5383 rabbits (the sum of AA and Aa genotypes) will be 99.685% or in parts it will be 0.99685.

Q 2 + 2q (1 - q) \u003d 0.99685 - this 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 = 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 = 0.056. And q then equals 0.944.

Q 2 is 0.891, and this is the proportion of homozygous chinchillas with the AA genotype. Since this value in% will be 89.1% of 5400 individuals, the number of homozygous chinchillas will be 4811 pcs.

Task 6. Determining the frequency of occurrence of heterozygous individuals from the known frequency of occurrence of recessive homozygotes

One form of glucosuria is inherited as an autosomal recessive trait and occurs at a frequency of 7:1,000,000. Determine the frequency of occurrence of heterozygotes in the population.

Let us designate 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 will be denoted by A.

Healthy individuals in the human population have the AA and Aa genotypes; diseased individuals have the genotype only aa.

Let us denote the frequency of occurrence of the recessive allele a by the letter q, and the frequency of 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 = 0.00264575

Since p + q = 1, then p = 1 - q = 0.9973543, and p2 = 0.9947155

Now substituting the p and q values ​​into the formula:

P2AA + 2pqAa + q2aa = 1,

Let's find the frequency of occurrence of heterozygous individuals 2pq in the human population:

2pq \u003d 1 - p 2 - q 2 \u003d 1 - 0.9947155 - 0.000007 \u003d 0.0052775.

Problem 7. Like the previous problem, but about albinism

General albinism (milky white skin color, 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 allele frequency a will be the square root of this number, that is, 0.0071. The allele frequency A will be 1 - 0.0071 = 0.9929, and the frequency of healthy AA homozygotes will be 0.9859.

The frequency of all heterozygotes 2Aa \u003d 1 - (AA + aa) \u003d 0.014 or 1.4%.

Problem 8. Everything seems so easy when you know how to solve it.

The population of Europeans according to the system of Rh blood groups contains 85% of Rh-positive individuals. Determine the saturation of the population with a recessive allele.

We know that the allelic gene responsible for the manifestation of Rh positive blood is dominant R (we denote the frequency of its occurrence by the letter p), and Rh negative is recessive r (we denote its frequency by the letter q).

Since the task says that p 2 RR + 2pqRr accounts for 85% of people, then the share of Rh-negative q 2 rr phenotypes will account for 15% or their frequency of occurrence will be 0.15 of all people of the European population.

Then the frequency of occurrence of the allele r or “saturation of the population with a recessive allele” (indicated by the letter q) will be the square root of 0.15 = 0.39 or 39%.

Task 9. The main thing is to know what penetrance is

Congenital dislocation of the hip is inherited dominantly. The average penetrance is 25%. The disease occurs with a frequency of 6:10,000. Determine the number of homozygous individuals in the population according to the recessive trait.

Penetrance is a quantitative measure of the phenotypic variability in the expression of a gene.

Penetrance is measured as the percentage of individuals in which a given gene manifested itself in the phenotype to the total number of individuals in whose genotype 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 a 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, let's denote it A. This means that organisms that have this disease have the genotypes AA and Aa.

It is known that hip dislocation is phenotypically detected in 6 organisms from the entire population (10,000 examined), but this is only one fourth of all people who actually have the AA and Aa genotypes (since it is said that the penetrance is 25%).

This means that in fact there are 4 times more people with AA and Aa genotypes, that is, 24 out of 10,000 or 0.0024 part. Then there will be 1 - 0.0024 = 0.9976 part of people with the aa genotype, or 9976 people out of 10,000.

Problem 10. If only men get sick

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 for the analyzed trait, 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%, 5 times more men are actually carriers of gout genes, 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 the AA + 2Aa genotypes in the population, which means that 95 out of 100 will be with the aa genotype.

If the frequency of occurrence of organisms with genotypes aa 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 = 0.975. Then the frequency of occurrence of the dominant allele "A" in this population is 1 - 0.975 = 0.005.

Target 11. How few people are resistant to HIV infection

Resistance to HIV infection is associated with the presence in the genotype of some recessive genes, such as CCR and SRF. The frequency of the recessive CCR-5 allele in the Russian population is 0.25%, and the frequency of 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 genotypes AA (homozygotes) or Aa (heterozygotes) are not resistant to HIV infection.

In the Russian population of resistant organisms for the CCR allelic gene there will be 0.25% squared = 0.0625%, and for the SRF allelic gene 0.05% squared = 0.0025%.

In the Kazakh population of resistant organisms for the CCR allelic gene there will be 0.12% squared = 0.0144%, and for the SRF allelic gene 0.1% squared = 0.01%.