3. POPULATION GENETICS.

Hardy Weinberg Law

Population  - this is a collection of individuals of one species, long occupying a certain range, freely interbreeding and relatively isolated from other individuals of the species.

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

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

Signs ideal population: population size  great exists panmixia  (there are no restrictions on the free choice of a partner), no mutations  on this basis, not valid natural selectionare absent inflowand   outflow of genes.

First position  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- frequency of the dominant allele A,q  - frequency of the recessive allele a. Both values \u200b\u200bare usually expressed in fractions of a unit, less often in percent (then p +q = 100 %).

Second position  Hardy-Weinberg Law: sum of genotype frequencies  one gene in a given population equal to one. The formula for calculating genotype frequencies is as follows:

p 2 + 2 pq + q 2 = 1 ,

where p2–   the frequency of homozygous individuals according to the dominant allele (genotype AA), 2pq -heterozygous frequency (genotype Aa), q2 –   the frequency of homozygous individuals according to the recessive allele (genotype aa).

The output of this formula is: in equilibrium population  females and males have the same frequencies as the A allele ( p) and allele a ( q) As a result of the crossing of female gametes ♀ ( p +q) with men ♂ ( p +q) and the frequencies of genotypes are determined: ( p +q) (p +q) = p2 + 2pq +q2 .

Third position  Law: in the equilibrium population allele frequenciesand   genotype frequencies persist for several generations.

TASKS

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

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

p2 + 2pq +q2 = 1,

where p- the frequency of the dominant gene, and q  - frequency of the recessive gene.

In this problem, the allele frequency A  equal to 0.8, and the frequency of the allele a equal to 0.2. Substituting these numerical values \u200b\u200bin the Hardy-Weinberg equation, we obtain the following expression:

0.82 + 2 × 0.8 × 0.2 + 0.22 \u003d 1 or 0.64 + 0.32 + 0.04 \u003d 1

From the equation it follows 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 - the frequency of the heterozygous genotype ( Aa).

3.2.   In the population of foxes per 1000 red, 10 white individuals are found. Determine the percentage of red homozygous, red heterozygous, and white foxes in this population.

Decision.

According to the equation:

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

3.3.   The brown-eyed allele dominates the blue-eyed. In the population, both alleles are found with equal probability.

Father and mother are brown-eyed. With what probability should we expect that the child born to them will be blue-eyed?

Decision.  If in the population both alleles are found equally often, then it contains 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 are against one, that this is a heterozygous. So, the probability of being heterozygous 2/3. The probability of transmitting the blue-eyed allele to offspring is 0 if the body is homozygous, and 1/2 if it is heterozygous. The full likelihood that this brown-eyed parent will transmit to the offspring an allele of blue eyes is 2 / 3x1 / 2, i.e. 1/3. For the child to be blue-eyed, he must receive from each of the parents an allele of blue eyes. This will happen with a probability of 1 / 3x1 / 3 \u003d 1/9.

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

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

p2 + 2pq +q2 = 1,

where p2 - the frequency of the dominant homozygous genotype, 2 pq  - the frequency of the heterozygous genotype and q2 - frequency of recessive homozygous genotype.

Cystic fibrosis of the pancreas affects individuals with a recessive homozygous phenotype; Consequently, q2 = 1 in 2000, or 1/2000 \u003d 0.0005. From here

Because the, p +q \u003d1, p \u003d1 – q \u003d1 – 0,0224 = 0,9776.

Thus, the frequency of the heterozygous phenotype (2 pq) \u003d 2 × (0.9776) × (0.0224) \u003d 0.044 \u003d 1 by 23.5%, that is, carriers of the recessive gene for pancreatic cystic fibrosis account for about 5% of the population.

3.5. A survey of the city N population (100,000 people) revealed 5 albinos. To establish the frequency of occurrence of heterozygous carriers of the albinism gene.

Decision.  Since albinos are recessive homozygotes ( aa), then, according to Hardy-Weinberg's law:   recessive gene frequency p +q \u003d1, from here, p \u003d1 – q; The frequency of heterozygotes is 2 pq.

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

3.6.   In a population of 5000 people, 64% are able to coagulate the tongue (dominant gene R), and 36% do not have this ability (recessive gene r) Calculate the frequency of occurrence of genes R  and r  and genotypes Rr, Rr  and rr  in a population.

Decision.  Frequency of occurrence of individuals with genotypes Rr  and Rr  equal to 0.64, and homozygotes rr (q2)   \u003d 0.36. Based on this, the frequency of the gene r  is equal to. And since p +q  \u003d 1 then p = 1 – q  \u003d 0.4, i.e., the allele frequency R (p)  is 0.4. If p\u003d 0.4, then p2 \u003d 0.16. Hence, the frequency of occurrence of individuals with a genotype Rr  is 16%.

So, the frequencies 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.   Three genotypes for the albinism gene are found in the population a  in the ratio: 9/16 AA, 6/16 Aa  and 1/16 aa. Is this population in a state of genetic equilibrium?

Decision.  It is known that the population consists of 9/16 AA, 6/16 Aa  and 1/16 aa  genotypes. Does this 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 a given attribute, is in a state of equilibrium: (3/4) 2 AA  : 2 × 3/4 × 1/4 Aa: (1/4)2 aa. From here

3.8.   Diabetes mellitus occurs in a population with a frequency of 1 in 200. Calculate the frequency of heterozygous carriers.

3.9.   Sickle cell anemia occurs in a population of people 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 part 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 Aa : 1296 aa?

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

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

3.15. Rye albinism is inherited as an autosomal recessive trait. In the surveyed area 84,000 plants. 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 fully dominate the white. Hybrids from crossing red with white have a roan suit. The Shorthorn breeding area has 4,169 red animals, 3,780 roan and 756 white animals.

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

3.17.*   One grain of wheat heterozygous for a certain gene accidentally fell on a desert island A. It ascended and gave rise to a series of generations that multiplied by self-pollination. What will be the proportions of heterozygous plants among representatives of the second, third, fourth, ..., n-th generations if, under the conditions controlled by the gene under consideration, under these conditions there is no effect on plant survival and the ability to produce offspring?

3.18.*   Snyder examined 3643 people on the ability to taste phenylthiourea and found that 70.2% of them are “sensory”, and 29.8% are “non-sensory”, this taste.

a) What is the proportion of “not feeling” children in marriages of “feeling” with “feeling”?

b) What is the proportion of children who “do not feel” the taste of phenylthiourea in marriages of “people who feel” and “do not feel” this taste?

Hardy-Weinberg Law

Population genetics deals with the genetic structure of populations.

The term “population” refers to a collection of freely interbreeding individuals of the same species, which has long existed in a certain territory (part of its range) and is relatively isolated from other populations of the same species.

The most important sign of a population is relatively free crossbreeding. If any insulating barriers arise that impede free crossbreeding, 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 significantly differ in genetic characteristics. In order to describe the genetic properties of a population, the concept of a gene pool is introduced: a set of genes found in a given population. In addition to the gene pool, the frequency of the gene or the frequency of the allele are also important.

Knowing how inheritance laws are implemented at the population level is crucial to understanding the causes of individual variability. All patterns identified in the course of psychogenetic studies relate to specific populations. In other populations, with a different gene pool and other gene frequencies, different results can be obtained.

Hardy-Weinberg law is the basis of mathematical constructions of population genetics and modern evolutionary theory. It was formulated independently by 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 quite large (ideally, infinitely large),

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

3) there is no mutation process,

4) there is no gene exchange with other populations,

5) natural selection is absent, 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. (Since the above conditions for the implementation 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, take the simplest case — the distribution of two alleles of the same gene. Let two organisms be the founders of a new population. One of them is dominant homozygote (AA), and the other is recessive homozygote (aa). Naturally, all their offspring in F 1 will be uniform and will have a genotype (Aa). Further, individuals F 1 will interbreed. We 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 unity, that is, p + q \u003d 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 be q. After similar reasoning for all spermatozoa in the population, we conclude that the proportion of spermatozoa carrying the allele (A) will be p, and those carrying the recessive allele (a) - q. Now we make up the Pennet 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 carried by them. At the intersection of rows and columns of the lattice, we obtain the descendants genotypes with coefficients corresponding to the frequencies 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 of recessive homozygotes (aa) is q. Since the given genotypes represent all possible variants of genotypes for the case under consideration, the sum of their frequencies should be equal to unity, i.e.

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

Consider an example of the use of this law in genetic calculations. It is known that one person out of 10 thousand is an albino, while the sign of albinism in a person is determined by one recessive gene. Let's calculate 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, one can determine the frequency of the albinism allele q, the frequency of the dominant allele of normal pigmentation p, and the frequency of the heterozygous genotype (2pq). People with this genotype will be the hidden carriers of albinism, despite the fact that this gene will not appear phenotypically in them and they will have normal skin pigmentation.

It can be seen from the simple calculations 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 phenotypically manifests itself very rarely, a significant number of carriers of this trait are present in the population, i.e., individuals having this gene in a heterozygote.

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

EXAMPLES OF SOLVING TASKS

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

Decision

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 - aah.Their frequency in the population (q 2 ) equal to 210/84000 \u003d 0.0025. Recessive gene frequency awill be equal to 0.0025. Consequently, q = 0,05.

Answer:0,05

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

Decision.

If the gene of the red suit of animals is denoted by A,
and the white gene a,then in red animals the genotype will be AA  (4169) Aa(3780), for whites - aa(756). In total, 8705 animals were registered. The frequency of homozygous red and white animals in units of one can be calculated. The frequency of white animals will be 756: 8705 \u003d 0.09. Therefore q 2 =0.09 . Recessive gene frequency q= = 0.3. Gene frequency Awill be p \u003d1 - q.Therefore r= 1 - 0,3 = 0,7.

Answer:  R\u003d 0.7, gene q \u003d0,3.

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.

Decision.

Albinism is inherited recessively. Value 1/20000 -
this is q 2 . Therefore, the frequency of the gene awill be: q \u003d1/20000 =
\u003d 1/141. The frequency of the p gene will be: r= 1 - q; R= 1 - 1/141 = 140/141.

The number of heterozygotes in a population is 2 pq\u003d 2 x (140/141) x (1/141) \u003d 1/70. Because in a population of 20,000 people, the number of heterozygotes in it is 1/70 x 20,000 \u003d 286 people.

Answer: 286 people

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

Decision.

Genotypes of people with congenital hip dislocation, AAand Aa(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 + 2 pq). However, the number of patients reported in the task 6/10000 represents only one fourth (25%) of gene A carriers in the population. Consequently, r 2   + 2pq \u003d(4 x 6) / 10,000 \u003d 24/10000. Then q 2 (the number of individuals homozygous for the recessive gene) is 1 - (24/10000) \u003d 9976/10000 or 9976 people.

Answer: 9976 people

4. In the population, allele frequencies p \u003d 0.8 and g \u003d 0.2 are known. Determine genotype frequencies.

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

5. The population has the following composition: 0.2AA, 0,3 Aa  and 0.50aa. Find allele frequenciesA  anda.

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

6. 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: heterozygous 42%; recessive homozygotes - 49%; dominant homozygotes - 9%.

7. Calculate genotype frequenciesAA, Aa  andaa  (in%) if individualsaa  make up 1% in the population.

Answer: in a population of 81% of individuals with a genotype AA, 18% with genotp Aa  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 non-eared (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.

We denote the frequency of occurrence of allele A in all gametes of the sheep population by the letter p, and the frequency of occurrence of allele a by the letter q. 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 q 2 is equal to 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) .

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: 1,000,000. Determine the frequency of occurrence of heterozygotes in the population.

2. 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.

3. 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.

4. The population of Europeans according to the system of blood groups Rhesus contains 85% Rhesus positive individuals. Determine the saturation of a 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 by the analyzed characteristic, based on these data.

Solution 1  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,
we find the frequency of occurrence of heterozygous 2pq individuals 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 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 2Aa \u003d 1 - (AA + aa) \u003d 0.014 or 1.4% .

Solution 3We 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 - this is the frequency of occurrence of albino rabbits with genotype aa. 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 equal to 0.891, and this is the share 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 .

Solution 4  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%.

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 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 .

The basis of population genetics, studying the genotypes of living organisms without the use of crosses, is the Hardy-Weinberg law (it was formulated independently by two scientists in 1908). This law (position, equilibrium, equation) is fully satisfied only under certain ideal conditions.

The Hardy-Weinberg equation is a mathematical model that explains how genetic balance is maintained in a population’s gene pool.

Hardy-Weinberg wording

The frequency of genotypes for a specific gene in a population remains constant over a series of generations and corresponds to the equation p 2 + 2pq + q 2 \u003d 1 (provided that there are only two alleles of this gene), where p 2 is the frequency (fraction of unit) of homozygotes for one allele ( for example, to the dominant one is AA), q 2 is the frequency of homozygotes for another allele (aa), 2pq is the frequency of heterozygotes (Aa), p is the proportion of the dominant allele in the population (A), q is the frequency of the recessive allele (a). Moreover, p + q \u003d 1, or A + a \u003d 1.

The formulation may be found not with respect to genotypes, but with respect to alleles: the frequencies of dominant and recessive alleles in a population will remain constant over a series of generations under a number of conditions. That is, the values \u200b\u200bof p and q will not change from generation to generation.

Thus, hardy-Weinberg law allows you to calculate the frequency of alleles and genotypes in a population, which is its important characteristic, since it is the population that is considered as a unit of evolution.

The frequencies of genotypes are calculated based on the frequencies of alleles and the probability of their association in one zygote. To calculate the frequencies of genotypes, the formula of the square of the binomial is used: (p + q) 2 \u003d p 2 + 2pq + q 2.

  Hardy-Weinberg Law Terms

Hardy-Weinberg's law is fully implemented when the following conditions are met:

The population must be large.

Individuals should not choose a mating partner, depending on the genotype of the studied genes. That is, mating should occur randomly.

The migration of individuals from and into the population should be absent.

In relation to the studied gene (its alleles), natural selection should not act. That is, all genotypes must be equally prolific.

No new mutations of the studied genes should arise.

For example, if in a population homozygotes for a recessive allele have reduced viability or are not selected by mating partners, then the Hardy-Weinberg law does not work for such a gene.

Thus, the frequency of alleles in a population remains constant if the crossing of individuals is random, and no external factors act on the population.

Deviation from the Hardy-Weinberg equation suggests that a certain evolutionary factor acts on the population. However, in large populations, deviations are insignificant when considering a short-term period of time, and this allows the use of the law for calculations. But on an evolutionary scale, the dynamics of a population’s gene pool reflects how evolution proceeds at the genetic level.

  Application of Hardy-Weinberg Law

In most cases, the frequency of alleles and genotypes is calculated based on the frequency of homozygous individuals according to the recessive allele. This is the only genotype that is recognized by phenotypic expression. It is often not possible to distinguish dominant homozygotes from heterozygotes; therefore, they are calculated using the Hardy-Weinberg equation.

Consider an example of using the Hardy-Weinberg law. Suppose that in a hypothetical population of people there are only two alleles of eye color - brown and blue. Brown color is determined by the dominant (A) allele of the gene, blue - recessive (a). Let the brown-eyed people be 75% (or in fractions of 0.75), and the blue-eyed people 25% (or 0.25). It is required to determine 1) the proportion of heterozygotes and dominant homozygotes, as well as 2) the frequency of alleles in the population.

If the proportion of recessive homozygotes is 0.25, then the proportion of the recessive allele is found as the square root of this number (based on the formula p 2 + 2pq + q 2 \u003d 1, where q 2 is the frequency of recessive homozygotes, and q is the frequency of the recessive allele), i.e. there will be 0.5 (or 50%). Since there are only two alleles in a hypothetical population of people, the sum of their shares will be one: p + q \u003d 1. From here we find the share of the dominant allele: p \u003d 1 - 0.5 \u003d 0.5. Thus, the frequency of both alleles is 50% each. We answered the second question.

The frequency of heterozygotes is 2pq. In this case, 2 * 0.5 * 0.5 \u003d 0.5. It follows that of 75% of brown-eyed people, 50% are heterozygotes. Then the share of dominant homozygotes remains 25%. We answered the first question of the problem.

Another example of the application of Hardy-Weinberg law. Such a human disease as cystic fibrosis occurs only in recessive homozygotes. The frequency of the disease is approximately 1 patient per 2500 people. This means that 4 out of 10,000 people are homozygotes, which amounts to 0.0004 in fractions of a unit. Thus q 2 \u003d 0.0004. Extracting the square root, we find the frequency of the recessive allele: q \u003d 0.02 (or 2%). The frequency of the dominant allele will be equal to p \u003d 1 - 0.02 \u003d 0.98. The frequency of heterozygotes: 2pq \u003d 2 · 0.98 · 0.02 \u003d 0.039 (or 3.9%). The value of the frequency of heterozygotes makes it possible to estimate the number of pathogenic genes that are in a latent state.

Such calculations show that, despite the small number of individuals with a homozygous recessive genotype, the frequency of the recessive allele in populations is quite high due to its presence in the genotypes of heterozygotes (carriers).

  In the case of multiple allelism

Hardy-Weinberg law applies to cases of multiple allelism. In this case, the same formula is used, but a polynomial from the frequencies of alleles is squared to determine the frequencies of genotypes.

If there are three alleles of a gene (a 1, a 2, a 3), then their frequencies (p, q, r) will add up to one: p + q + r \u003d 1. If you square the equation, we get the following distribution genotype frequencies:

(p + q + r) 2 \u003d p 2 + q 2 + r 2 + 2pq + 2pr + 2qr \u003d 1

Here p 2, q 2, r 2 are the frequencies of homozygotes (respectively a 1 a 1, a 2 a 2, a 3 a 3). And 2pq, 2pr, 2qr are the frequencies of heterozygous genotypes (a 1 a 2, a 1 a 3, a 2 a 3). The sum of the frequencies of genotypes, as well as the sum of the frequencies of alleles, will always be 1.

One of the most important applications of the Hardy-Weinberg law is that it makes it possible to calculate some of the frequencies of genes and genotypes in the case when not all genotypes can be identified due to the dominance of some alleles.

Example 1: albinism in humans is due to a rare recessive gene. If the normal pigmentation allele is designated A, and the albinism allele is a, then the albino genotype will be aa, and the genotypes of normally pigmented people will be AA and Aa. Suppose that in a population of people (the European part) the frequency of albinos is 1 in 10,000. According to Hardy-Weinberg's law, in this population the frequency of homozygotes is q 2 aa \u003d 1: 10000 \u003d 0.0001 (0.1%), and the frequency of recessive homozygotes \u003d 0.01. The frequency of the dominant allele is pA \u003d 1-qa \u003d 1-0.01 \u003d 0.99. The frequency of normally pigmented people is p 2 AA \u003d 0.99 2 \u003d 0.98 (98%), and the frequency of heterozygotes 2pqAa \u003d 2 × 0.99 × 0.1 \u003d 0.198 (1.98%).

An important consequence of Hardy-Weinberg's law is that rare alleles are present in a population mainly in a heterozygous state. Consider the given example with albinism (genotype aa). The frequency of albinos is 0.0001, and the heterozygote Aa is 0.00198. The frequency of the recessive allele in heterozygotes is half the frequency of heterozygotes, i.e. 0.0099. Therefore, in the heterozygous state contains about 100 times more recessive alleles than in the homozygous state. Thus, the lower the frequency of the recessive allele, the greater the proportion of this allele present in the population in a heterozygous state.

Example 2: the frequency of phenylketonuria (PKU) in a population is 1: 10000, PKU is an autosomal recessive disease, therefore individuals with genotypes AA and Aa are healthy, with genotypes aa they are sick with PKU.

The population is therefore represented by genotypes in the following ratio:

p 2 AA + 2pqAa + q 2 aa \u003d 1

Based on these conditions:

q 2 aa \u003d 1/10000 \u003d 0.0001.

pA \u003d 1-qa \u003d 1-0.01 \u003d 0.99

p 2 AA \u003d 0.99 2 \u003d 0.9801

2paAa \u003d 2 × 0.99 × 0.01 \u003d 0.0198, or ~ 1.98% (2%)

Therefore, in this population, the frequency of heterozygotes for the PKU gene in the studied population is approximately 2%. The number of individuals with the AA genotype is 10,000 × 0.9801 \u003d 9801, the number of individuals with the Aa genotype (carriers) is 10,000 × 0.0198 \u003d 198 people, because the relative fractions of genotypes in this population are represented by the ratio 1 (aa): 198 (Aa): 980 (AA).

In the event that the gene in the gene pool is represented by several alleles, for example, the blood group I gene of the AB0 system, then the ratio of different genotypes is expressed by the formula (and the Hardy-Weinberg principle remains valid.

For example: among the Egyptians there are blood groups in the AB0 system in the following percentage ratio:

0 (I) - 27.3%, A (II) - 38.5%, B (III) - 25.5%, AB (IV) - 8.7%

Determine the frequency of alleles I 0, I A, I B and different genotypes in this population.

When solving the problem, you can use the formulas:

  ; (;, where A is the frequency of blood group A (II); 0 is the frequency of blood group 0 (I); B is the frequency of blood group B (III).

Verification: pI A + qI B + rI 0 \u003d 1 (0.52 + 0.28 + 0.20 \u003d 1).

For genes linked to sex, the equilibrium frequencies of X A 1 X A 1, X A 1 X A 2 and X A 2 X A 2 coincide with those for autosomal genes: p 2 + 2pq + q 2. For males (in the case of heterogametic sex), due to hemizygosity, only two genotypes X A 1 Y or X A 2 Y 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. From this it follows that phenotypes determined by recessive alleles linked to the X chromosome are more common in males than in females. So, with the frequency of the hemophilia allele qa \u003d 0.0001, in men the disease occurs 10,000 times more often than in women (1 / 10000mln in men and 1 / 100mln in women).

In order to establish and confirm the type of disease inheritance, it is necessary to check whether segregation in burdened families of a given population corresponds to Mendeleev's laws. The c-square method confirms the correspondence of the number of patients and healthy siblings for autosomal pathology in families with full registration (through sick parents).

A number of methods can be used to calculate segregation frequency: Weinberg sibling method, proband method.

Exercise 1.

Study the lecture notes and study material.

Task 2.

Write down in the dictionary and learn the basic terms and concepts: population, panmixia, panmix population, gene pool, allele frequency, phenotype and genotype frequency in a population, Hardy-Weinberger law (its content), genetic structure of a population, balance of the genetic structure of a population over generations, mutational pressure, genetic load, selection coefficient, population genetic analysis, factors of the genetic dynamics of a population, genetic drift, inbreeding, adaptation coefficient.

Task 3.

Model the panmix population and draw a conclusion about its genetic structure and genetic balance in a series of generations (on the instructions of the teacher), in two versions, with s \u003d 0 and s \u003d -1®aa.

Gametes are conventionally represented by cardboard circles. A dark circle indicates a gamete with a dominant allele. A, white - with a recessive allele a. Each subgroup receives two sacs, in which one hundred “gametes”: in one - “eggs”, in the other - “sperm”: for example, A - 30 circles, and - 70 circles, total - 100 spermatozoa and also eggs. One of the students gets, without looking, one circle (“eggs”), another similarly takes out circles - “sperm”, the third student writes the resulting genotype combination to Table 5 using the envelope rule. The combination of two dark circles means AA, homozygous for the dominant; two white aarecessive homozygote; dark and white - Aaheterozygote. Since the combination of circles – gametes is random, the process is simulated panmixia.

Table 5. The number of genotypes and the frequency of alleles in the model population

In the second option, work should be done until the number of genotypes is repeated, which indicates the establishment of a new equilibrium state in the population.

When recording genotypes, both random errors can crept in, and a regular change in the number of the genotype can be reflected. Therefore, it is necessary to calculate the criterion χ 2 -a criterion for the compliance of practically obtained data with the theoretically expected one.

To do this, we determine the theoretically expected genotype frequency for a given gamete ratio. For example, if the original gametes are: circles A – 30, a  –70; then according to the Pennet table:

χ 2 fact. \u003d Σd 2 / q \u003d 9: 9 + 36: 42 + 9: 49 \u003d 1 + 0.85 + 0.18 \u003d 2.03; when n "\u003d 2, when P \u003d 0.05

Comparing method χ 2  the results obtained with the theoretically expected results, we conclude that in this case the ratio obtained does not differ from the expected one, since χ 2 fact.< χ2 tabular 5.99. Therefore, in version I, the initial allele frequencies (pA - 03 and qa - 0.3) are preserved in the panmix population. Carry out similar work for options I and II. Draw conclusions.

Task 4.

Solve the following tasks:

1. Tay-Sachs disease   caused by an autosomal recessive allele. The characteristic signs of this disease are mental retardation and blindness, death occurs at the age of about four years. The frequency of illness among newborns is about ten per 1 million. Based on the Hardy-Weinberg balance, calculate the frequencies of alleles and heterozygotes.

2. Cystic fibrosis   pancreatic tissue ( cystic fibrosis ) - a hereditary disease caused by a recessive allele; characterized by poor absorption in the intestine and abstructive changes in the lungs and other organs. Death usually occurs at the age of about 20 years. Among newborns, cystic fibrosis occurs on average in 4 per 10,000. Based on the Hardy – Weinberg balance, calculate the frequencies of all three genotypes in newborns, what percentage heterozygous carriers are.

3. Akatalazia   - A disease caused by a recessive gene was first discovered in Japan. In heterozygotes for this gene, a decreased catalase level in the blood is observed. The heterozygous frequency is 0.09% among the population of Hiroshima and Nagasaki; and 1.4% of the rest of the Japanese population. Based on the Hardy – Weinberg equilibrium, calculate the frequencies of alleles and genotypes:

In Hiroshima and Nagasaki;

Among the rest of the population of Japan.

Task 4. The table shows the frequency of alleles that control blood groups of the AB0 system among people from 4 examined populations. Determine the frequency of different genotypes in each of these populations.

Table 6. The frequency of alleles that determine blood groups AB0

5. The table shows the frequency (in percent) of blood groups 0, A, B and AB in 4 different populations. Determine the frequency of the corresponding alleles and different genotypes in each of these populations.

Table 7. The frequency of blood groups AB0

Task 5.

Answer the self-test questions:

1. Explain what you need to understand by the genetic and genotypic structure of the population.

2. What law obeys the genetic structure of a population, what is its essence.

3. Describe the factors of dynamic processes in the population.

4. The selection coefficient, its essence.

5. Why are hereditary diseases more often manifested in closely related marriages?

6. What genotypes contain recessive alleles in populations.

Report Form:

Provision of a workbook for verification;

Solving problems to determine the genetic structure of a population using the Hardy-Weinberg Law;

Oral protection of work performed.

Different ways of expressing frequency calculation,

Pronounced allele frequency in fractions of a unit

Or genotype in a population

1. In the study population 84 people 84: 420 \u003d 0, 2

of 420 had a dominant trait.

2. In one of the populations, the incidence is 15: 100 \u003d 0.15

people with rhesus positive blood

(recessive symptom) is 15%.

3. The incidence of patients suffering from 10 -4 \u003d 1: 10000 \u003d 0.0001

phenylketonuria is equal to 10 -4.

4. In European populations, 0.02: 1000 \u003d 0.00002

prevalence of achondroplasia

is 0.02 per 1000 newborns.

5. Alcaptonuria occurs with a frequency of 1: 100 000 \u003d 0.00001

6. The studied sign is characterized by 0.09: 0.3 \u003d 0.3

incomplete penetrance equal to

30%, and occurs in a population with

frequency of 0.09.

Genotype frequency  - the proportion of individuals in a population having a given genotype among all individuals in a population.

Allele frequency  - the proportion of a specific allele among all the alleles in the population of the studied gene.

A Pair of Alternative Alleles Possible Genotypes

Gene traits

Albinism a (q) aa (q 2)

Lack of Albinism A (r) A _(p 2 + 2 pq): AA (p 2) or Aa (2pq)

The frequency of homozygotes for a recessive trait in a population:

q 2 \u003d 1: 20,000 \u003d 0.00005

The frequency of a recessive allele in a population:

The frequency of the dominant allele in a population:

p \u003d 1 - q \u003d 1–0.07 \u003d 0.93

The frequency of heterozygotes in the population:

2pq \u003d 2 * 0.07 * 0.93 \u003d 0.1302 (13%)

Answer:The frequency of heterozygotes in the population is 13%.

1. One of the forms of fructosuria (weakening the absorption of fructose and its increased content in the sword) is subclinical. Exchange defects are reduced with the exclusion of fructose from food. The disease is inherited autosomally recessively and occurs with a frequency of 7: 1,000,000 (V.P. Efroimson, 1968) Determine the frequency of heterozygotes in the population.

2. Congenital hip dislocation is inherited dominantly, the average penetrance of the gene is 25%. The disease occurs with a frequency of 0.06% (V.P. Efroimson, 1968). Determine the number of homozygous individuals by the recessive gene.

3. In one panmictic population, the frequency of alleles b is 0.1, and in the other, 0.9. Which population has more heterozygotes?

4. 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. How many healthy people will live in a population of 400,000?

5. Cystic fibrosis of the pancreas (cystic fibrosis) affects individuals with a recessive homozygous phenotype and is found in a population with a frequency of 1 in 2000. Calculate the frequency of the cystic fibrosis gene in a population of 1,000,000.

6. In the population, there are three genotypes for the eye color gene in the ratio: 9 / 16AA, 6 / 16Aa and 1 / 16aa. Brown eye color is an autosomal dominant trait with constant penetrance. Is this population in a state of genetic equilibrium?

7. Aniridia is inherited as a dominant autosomal trait and occurs with a frequency of 1: 10000 (VP Efroimson). Determine the genetic structure of the population.

8. Gangington's chorea is inherited as an autosomal dominant trait with a penetrance of 82.5%. In the population per 100 thousand people there are 4 patients. Determine the percentage of people carrying the disease in the population.

9. The population frequency of craniofacial dysostosis is 1:25 000. This trait is inherited autosomally dominantly with a penetrance of 50%. How many people in the population will be carriers of this gene.

10. 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.

11. From the diseases listed below, indicate those whose population numbers can be calculated using the Hardy-Weinberg law: Patau syndrome, Jacob syndrome, phenylketonuria, polydactyly, sickle cell anemia, cat scream syndrome, hypertrichosis, color blindness.

12. Tuberous sclerosis (epiloy) is inherited as an autosomal dominant trait. According to Penrose (1972), this disease occurs with a frequency of 1: 600,000. One of the symptoms of this disease - fundus phacoma (retinal tumor) - is found in 80% of all homozygotes and 20% of presumably heterozygous, which have no other clinical symptoms . Determine the frequency of occurrence of the dominant gene (solving the problem at the request of the student).