For this task, you can get 3 points on the exam in 2020

The topic of task 28 of the Unified State Examination in biology was "Supraorganismal systems and the evolution of the world." Many schoolchildren note the complexity of this test due to the large amount of educational material it covers, as well as due to the construction of the ticket. In task No. 28, the compiler, the Russian FIPI, the Federal Institute of Pedagogical Measurements, offers six options for each question, the correct of which may be any number from one to all six. Sometimes the question itself contains a hint - how many options you will have to choose (“Which three signs of the listed six are characteristic of animal cells”), but in most cases the student himself must decide on the number of answers he chooses as correct.

The questions of task 28 of the USE in biology may also affect the basics of biology. Be sure to repeat before the exams - what is the absence of artificial and natural ecosystems, water and terrestrial, meadow and field, how the rule of the ecological pyramid sounds and where it applies, what is biogeocenosis and agrocenosis. Some questions are logical in nature, you need to not only rely on the theory from a school textbook, but also think logically: “In a mixed forest, plants are arranged in tiers, and this is the reason for the decrease in competition between a birch and another living organism. Which one? May beetle, bird cherry, mushrooms, wild rose, hazel, mice are offered as answers. In this case, the student must remember that competition always goes for the same resources, in this case (with a tiered arrangement of plants) for light, so only trees and shrubs need to be selected from the list - bird cherry, wild rose and hazel.

Among the tasks in genetics at the exam in biology, 6 main types can be distinguished. The first two - to determine the number of types of gametes and monohybrid crossing - are most often found in part A of the exam (questions A7, A8 and A30).

Tasks of types 3, 4 and 5 are devoted to dihybrid crossing, inheritance of blood groups and sex-linked traits. Such tasks make up the majority of C6 questions in the exam.

The sixth type of tasks is mixed. They consider the inheritance of two pairs of traits: one pair is linked to the X chromosome (or determines human blood groups), and the genes of the second pair of traits are located on autosomes. This class of tasks is considered the most difficult for applicants.

This article sets out theoretical foundations of genetics necessary for successful preparation for task C6, as well as solutions to problems of all types are considered and examples for independent work are given.

Basic terms of genetics

Gene- This is a section of the DNA molecule that carries information about the primary structure of one protein. A gene is a structural and functional unit of heredity.

Allelic genes (alleles)- different variants of the same gene encoding an alternative manifestation of the same trait. Alternative signs - signs that cannot be in the body at the same time.

Homozygous organism- an organism that does not give splitting for one reason or another. Its allelic genes equally affect the development of this trait.

heterozygous organism- an organism that gives splitting according to one or another feature. Its allelic genes affect the development of this trait in different ways.

dominant gene is responsible for the development of a trait that manifests itself in a heterozygous organism.

recessive gene is responsible for the trait, the development of which is suppressed by the dominant gene. A recessive trait appears in a homozygous organism containing two recessive genes.

Genotype- a set of genes in the diploid set of an organism. The set of genes in a haploid set of chromosomes is called genome.

Phenotype- the totality of all the characteristics of an organism.

G. Mendel's laws

Mendel's first law - the law of uniformity of hybrids

This law is derived on the basis of the results of monohybrid crossing. For experiments, two varieties of peas were taken, differing from each other in one pair of traits - the color of the seeds: one variety had a yellow color, the second - green. Crossed plants were homozygous.

To record the results of crossing, Mendel proposed the following scheme:

Yellow seed color
- green seed color

(parents)
(gametes)
(first generation)	(all plants had yellow seeds)

The wording of the law: when crossing organisms that differ in one pair of alternative traits, the first generation is uniform in phenotype and genotype.

Mendel's second law - the law of splitting

Plants were grown from seeds obtained by crossing a homozygous plant with yellow seeds with a plant with green seeds, and by self-pollination was obtained.

The wording of the law: in the offspring obtained from crossing hybrids of the first generation, there is a splitting according to the phenotype in the ratio, and according to the genotype -.

Mendel's third law - the law of independent inheritance

This law was derived on the basis of data obtained during dihybrid crossing. Mendel considered the inheritance of two pairs of traits in peas: seed color and shape.

As parental forms, Mendel used plants homozygous for both pairs of traits: one variety had yellow seeds with a smooth skin, the other green and wrinkled.

Yellow seed color - green color of seeds,
- smooth shape, - wrinkled shape.

Then Mendel grew plants from seeds and obtained second-generation hybrids by self-pollination.

In there was a splitting into phenotypic class in the ratio . all seeds had both dominant traits (yellow and smooth), - the first dominant and the second recessive (yellow and wrinkled), - the first recessive and the second dominant (green and smooth), - both recessive traits (green and wrinkled).

When analyzing the inheritance of each pair of traits, the following results are obtained. In parts of yellow seeds and parts of green seeds, i.e. ratio . Exactly the same ratio will be for the second pair of characters (seed shape).

The wording of the law: when crossing organisms that differ from each other by two or more pairs of alternative traits, genes and their corresponding traits are inherited independently of each other and combined in all possible combinations.

Mendel's third law holds only if the genes are on different pairs of homologous chromosomes.

Law (hypothesis) of "purity" of gametes

When analyzing the characteristics of hybrids of the first and second generations, Mendel found that the recessive gene does not disappear and does not mix with the dominant one. In both genes are manifested, which is possible only if the hybrids form two types of gametes: one carries a dominant gene, the other a recessive one. This phenomenon is called the gamete purity hypothesis: each gamete carries only one gene from each allelic pair. The hypothesis of gamete purity was proved after studying the processes occurring in meiosis.

The hypothesis of "purity" of gametes is the cytological basis of Mendel's first and second laws. With its help, splitting by phenotype and genotype can be explained.

Analyzing cross

This method was proposed by Mendel to determine the genotypes of organisms with a dominant trait that have the same phenotype. To do this, they were crossed with homozygous recessive forms.

If, as a result of crossing, the entire generation turned out to be the same and similar to the analyzed organism, then it could be concluded that the original organism is homozygous for the trait under study.

If, as a result of crossing, a splitting in the ratio was observed in the generation, then the original organism contains the genes in a heterozygous state.

Inheritance of blood groups (AB0 system)

The inheritance of blood groups in this system is an example of multiple allelism (the existence of more than two alleles of one gene in a species). There are three genes in the human population that code for erythrocyte antigen proteins that determine people's blood types. The genotype of each person contains only two genes that determine his blood type: the first group; second and ; third and fourth.

Inheritance of sex-linked traits

In most organisms, sex is determined at the time of fertilization and depends on the set of chromosomes. This method is called chromosomal sex determination. Organisms with this type of sex determination have autosomes and sex chromosomes - and.

In mammals (including humans), the female sex has a set of sex chromosomes, the male sex -. The female sex is called homogametic (forms one type of gametes); and male - heterogametic (forms two types of gametes). In birds and butterflies, males are homogametic and females are heterogametic.

The USE includes tasks only for traits linked to the -chromosome. Basically, they relate to two signs of a person: blood clotting (- normal; - hemophilia), color vision (- normal, - color blindness). Tasks for the inheritance of sex-linked traits in birds are much less common.

In humans, the female sex may be homozygous or heterozygous for these genes. Consider the possible genetic sets in a woman on the example of hemophilia (a similar picture is observed with color blindness): - healthy; - healthy, but is a carrier; - sick. The male sex for these genes is homozygous, tk. - chromosome does not have alleles of these genes: - healthy; - is ill. Therefore, men are most often affected by these diseases, and women are their carriers.

Typical USE tasks in genetics

Determination of the number of types of gametes

The number of gamete types is determined by the formula: , where is the number of gene pairs in the heterozygous state. For example, an organism with a genotype has no genes in a heterozygous state; , therefore, and it forms one type of gamete. An organism with a genotype has one pair of genes in a heterozygous state, i.e. , therefore, and it forms two types of gametes. An organism with a genotype has three pairs of genes in a heterozygous state, i.e. , therefore, and it forms eight types of gametes.

Tasks for mono- and dihybrid crossing

For a monohybrid cross

Task: Crossed white rabbits with black rabbits (black color is a dominant trait). In white and black. Determine the genotypes of parents and offspring.

Decision: Since splitting is observed in the offspring according to the trait being studied, therefore, the parent with the dominant trait is heterozygous.

	(black)	(white)
	(black) : (white)

For a dihybrid cross

Dominant genes are known

Task: Crossed tomatoes of normal growth with red fruits with dwarf tomatoes with red fruits. All plants were of normal growth; - with red fruits and - with yellow ones. Determine the genotypes of parents and offspring if it is known that in tomatoes the red color of the fruit dominates over yellow, and normal growth over dwarfism.

Decision: Denote dominant and recessive genes: - normal growth, - dwarfism; - red fruits, - yellow fruits.

Let us analyze the inheritance of each trait separately. All offspring have normal growth, i.e. splitting on this basis is not observed, so the original forms are homozygous. Splitting is observed in fruit color, so the original forms are heterozygous.

Dominant genes unknown

Task: Two varieties of phlox were crossed: one has red saucer-shaped flowers, the second has red funnel-shaped flowers. The offspring produced red saucers, red funnels, white saucers and white funnels. Determine the dominant genes and genotypes of parental forms, as well as their descendants.

Decision: Let us analyze the splitting for each feature separately. Among the descendants, plants with red flowers are, with white flowers -, i.e. . Therefore, red - white color, and parental forms are heterozygous for this trait (because there is splitting in the offspring).

Splitting is also observed in the shape of the flower: half of the offspring have saucer-shaped flowers, half are funnel-shaped. Based on these data, it is not possible to unambiguously determine the dominant trait. Therefore, we accept that - saucer-shaped flowers, - funnel-shaped flowers.

red saucer-shaped flowers,
- red funnel-shaped flowers,
- white saucer-shaped flowers,
- white funnel-shaped flowers.

Solving problems on blood groups (AB0 system)

Task: the mother has the second blood group (she is heterozygous), the father has the fourth. What blood groups are possible in children?

Decision:

Solving problems on the inheritance of sex-linked traits

Such tasks may well occur both in part A and in part C of the USE.

Task: a carrier of hemophilia married a healthy man. What kind of children can be born?

Decision:

Solving problems of mixed type

Task: A man with brown eyes and blood type marries a woman with brown eyes and blood type. They had a blue-eyed child with a blood type. Determine the genotypes of all individuals indicated in the problem.

Decision: Brown eye color dominates blue, therefore - brown eyes, - Blue eyes. The child has blue eyes, so his father and mother are heterozygous for this trait. The third blood group may have the genotype or, the first - only. Since the child has the first blood type, therefore, he received the gene from both his father and mother, therefore his father has a genotype.

Task: The man is colorblind, right-handed (his mother was left-handed), married to a woman with normal vision (her father and mother were completely healthy), left-handed. What kind of children can this couple have?

Decision: In a person, the best possession of the right hand dominates over left-handedness, therefore - right-handed, - lefty. Male genotype (because he received the gene from a left-handed mother), and women -.

A color-blind man has the genotype, and his wife -, because. her parents were completely healthy.

Tasks for independent solution

1. Determine the number of types of gametes in an organism with a genotype.
2. Determine the number of types of gametes in an organism with a genotype.
3. They crossed tall plants with short plants. B - all plants are medium in size. What will be?
4. They crossed a white rabbit with a black rabbit. All rabbits are black. What will be?
5. They crossed two rabbits with gray wool. B with black wool, - with gray and white. Determine the genotypes and explain this splitting.
6. They crossed a black hornless bull with a white horned cow. They received black hornless, black horned, white horned and white hornless. Explain this split if black and the absence of horns are dominant traits.
7. Drosophila with red eyes and normal wings were crossed with fruit flies with white eyes and defective wings. The offspring are all flies with red eyes and defective wings. What will be the offspring from crossing these flies with both parents?
8. A blue-eyed brunette married a brown-eyed blonde. What kind of children can be born if both parents are heterozygous?
9. A right-handed man with a positive Rh factor married a left-handed woman with a negative Rh factor. What kind of children can be born if a man is heterozygous only for the second sign?
10. The mother and father have a blood type (both parents are heterozygous). What blood group is possible in children?
11. The mother has a blood group, the child has a blood group. What blood type is impossible for a father?
12. The father has the first blood type, the mother has the second. What is the probability of having a child with the first blood group?
13. A blue-eyed woman with a blood type (her parents had a third blood type) married a brown-eyed man with a blood type (his father had blue eyes and a first blood type). What kind of children can be born?
14. A right-handed hemophilic man (his mother was left-handed) married a left-handed woman with normal blood (her father and mother were healthy). What kind of children can be born from this marriage?
15. Strawberry plants with red fruits and long-leaved leaves were crossed with strawberry plants with white fruits and short-leaved leaves. What kind of offspring can there be if red color and short-leaved leaves dominate, while both parental plants are heterozygous?
16. A man with brown eyes and blood type marries a woman with brown eyes and blood type. They had a blue-eyed child with a blood type. Determine the genotypes of all individuals indicated in the problem.
17. They crossed melons with white oval fruits with plants that had white spherical fruits. The following plants were obtained in the offspring: with white oval, with white spherical, with yellow oval and with yellow spherical fruits. Determine the genotypes of the original plants and descendants, if the white color of the melon dominates over the yellow, the oval shape of the fruit is over the spherical.

Answers

1. gamete type.
2. gamete types.
3. gamete type.
4. high, medium and low (incomplete dominance).
5. black and white.
6. - black, - white, - grey. incomplete dominance.
7. Bull:, cow -. Offspring: (black hornless), (black horned), (white horned), (white hornless).
8. - Red eyes, - white eyes; - defective wings, - normal. Initial forms - and, offspring.
   Crossing results:
   a)
9. - Brown eyes, - blue; - dark hair, - light. Father mother - .
   

   	- brown eyes, dark hair - brown eyes, blonde hair - blue eyes, dark hair - blue eyes, blonde hair

10. - right-handed, - left-handed; Rh positive, Rh negative. Father mother - . Children: (right-handed, Rh positive) and (right-handed, Rh negative).
11. Father and mother - . In children, a third blood type (probability of birth -) or a first blood group (probability of birth -) is possible.
12. Mother, child; He received the gene from his mother, and from his father -. The following blood types are impossible for the father: second, third, first, fourth.
13. A child with the first blood group can only be born if his mother is heterozygous. In this case, the probability of birth is .
14. - Brown eyes, - blue. Female Male . Children: (brown eyes, fourth group), (brown eyes, third group), (blue eyes, fourth group), (blue eyes, third group).
15. - right-handed, - lefty. Man Woman . Children (healthy boy, right-handed), (healthy girl, carrier, right-handed), (healthy boy, left-handed), (healthy girl, carrier, left-hander).
16. - red fruit - white; - short-stalked, - long-stalked.
    Parents: and Offspring: (red fruit, short stem), (red fruit, long stem), (white fruit, short stem), (white fruit, long stem).
    Strawberry plants with red fruits and long-leaved leaves were crossed with strawberry plants with white fruits and short-leaved leaves. What offspring can there be if red color and short-leaved leaves dominate, while both parental plants are heterozygous?
17. - Brown eyes, - blue. Female Male . Child:
18. - white color, - yellow; - oval fruits, - round. Source plants: and. Offspring:
    with white oval fruits,
    with white spherical fruits,
    with yellow oval fruits,
    with yellow spherical fruits.

The task belongs to the highest level of difficulty. For the correct answer you will receive 3 points.

Approximately up to 10-20 minutes.

To complete task 28 in biology, you need to know:

• how (compose crossbreeding schemes), ecology, evolution;

Tasks for training

Task #1

The hamster color gene is linked to the X chromosome. The X A genome is determined by brown color, the X B genome is black. Heterozygotes are tortoiseshell. Five black hamsters were born from a tortoiseshell female and a black male. Determine the genotypes of parents and offspring, as well as the nature of the inheritance of traits.

Task #2

In Drosophila, the black color of the body dominates over the gray, normal wings - over the curved ones. Two black flies with normal wings are crossed. The offspring of F 1 are phenotypically uniform - with a black body and normal wings. What are the possible genotypes of the crossed individuals and offspring?

Task #3

A person has four phenotypes according to blood groups: I (0), II (A), III (B), IV (AB). The gene that determines the blood group has three alleles: I A , I B , i 0 , and the i 0 allele is recessive with respect to the IA and IB alleles. The gene for color blindness d is linked to the X chromosome. A woman with blood group II (heterozygote) and a man with blood group III (homozygote) entered into marriage. It is known that the woman's father suffered from color blindness, the mother was healthy. The man's relatives never had this disease. Determine the genotypes of the parents. Specify the possible genotypes and phenotypes (blood group number) of children. Make a scheme for solving the problem. Determine the probability of the birth of children with color blindness and children with II blood group.

Task #4

In corn, the genes for brown color and smooth seed shape dominate over the genes for white color and wrinkled shape.

When plants with brown smooth seeds were crossed with plants with white color and wrinkled seeds, 4006 seeds of brown smooth and 3990 seeds of white wrinkled, as well as 289 white smooth and 316 brown wrinkled seeds of corn were obtained. Make a scheme for solving the problem. Determine the genotypes of maize parent plants and its offspring. Justify the appearance of two groups of individuals with traits different from their parents.

Secondary general education

Line UMK VV Pasechnik. Biology (10-11) (base)

Line UMK Ponomareva. Biology (10-11) (B)

Biology

USE in biology-2018: task 27, basic level

Task 27: what's new?

Part of the tasks of this line has changed: now it is more often required to predict the consequences of a mutation of a gene section. First of all, there will be variants of tasks for gene mutations, but it is also appropriate to repeat chromosomal and genomic mutations.

In general, task number 27 this year is represented by very diverse options. Some of the tasks are related to protein synthesis. It is important to understand here: the algorithm for solving a problem depends on how it is formulated. If the task begins with the words “it is known that all types of RNA are transcribed into DNA” - this is one synthesis sequence, but it can be proposed to simply synthesize a polypeptide fragment. Regardless of the wording, it is extremely important to remind students how to write DNA nucleotide sequences correctly: without spaces, hyphens and commas, with a continuous sequence of characters.

To correctly solve problems, you need to carefully read the question, paying attention to additional comments. The question may sound like this: what changes can occur in a gene as a result of a mutation if one amino acid is replaced by another in a protein? What property of the genetic code determines the possibility of the existence of different fragments of a mutated DNA molecule? The task may also be given to restore a DNA fragment in accordance with the mutation.

If the problem contains the wording “explain using your knowledge of the properties of the genetic code”, it would be appropriate to list all the properties that are known to students: redundancy, degeneracy, non-overlapping, etc.

What topics should be studied in order to successfully solve the problems of the 27th line?

• Mitosis, meiosis, plant development cycles: algae, mosses, ferns, gymnosperms, angiosperms.


• Microsporogenesis and macrosporogenesis in gymnosperms and angiosperms.

A new study guide is offered to the attention of students and teachers, which will help them successfully prepare for the unified state exam in biology. The collection contains questions selected by sections and topics tested at the exam, and includes tasks of different types and levels of complexity. Answers to all questions are given at the end of the manual. The proposed thematic tasks will help the teacher organize preparation for the unified state exam, and students will independently test their knowledge and readiness for the final exam. The book is addressed to students, teachers and methodologists.

How to prepare?

1. Show students diagrams and algorithms: how plants form spores and gametes, how animals form gametes and somatic cells. It is useful to ask students to model their own mitosis and meiosis schemes: this allows you to understand why the haploid cells formed during meiosis later become diploid.


2. Turn on visual memory. It is useful to memorize illustrations of the basic schemes of the evolution of the life cycle of various plants - for example, the cycle of alternation of generations in algae, ferns, bryophytes. Unexpectedly, but questions related to the life cycle of pine, for some reason, often cause difficulties. The topic itself is not complicated: it is enough to know about microsporangia and megasporangia that they are formed by meiosis. It must be understood that the bump itself is diploid: for the teacher this is obvious, but for the student it is not always.


3. Pay attention to the nuances of the wording. When describing some issues, clarifications need to be made: in the life cycle of brown algae, an alternation of a haploid gametophyte and a diploid sporophyte is observed, with the latter predominating (this way we will get rid of possible nit-picking). A nuance in the topic of the life cycle of ferns: explaining from what and how spores are formed, one can answer in different ways. One option is from sporagon cells, and the other, more convenient, is from spore mother cells. Both answers are satisfactory.

We analyze examples of tasks

Example 1 A fragment of a DNA chain has the following sequence: TTTGCGATGCCCCA. Determine the sequence of amino acids in the polypeptide and justify your answer. What changes can occur in a gene as a result of a mutation in if the third amino acid in the protein has been replaced by the amino acid CIS? What property of the genetic code determines the possibility of the existence of different fragments of a mutated DNA molecule? Explain your answer using the genetic code table.

Decision. This task is easily decomposed into elements that will make up the correct answer. It is best to act according to a proven algorithm:

1. determine the sequence of amino acids in the fragment;


2. write what happens when one amino acid is replaced;


3. we conclude that there is a degeneracy of the genetic code: one amino acid is encoded by more than one triplet (here, the skill of solving such problems is needed).

Example 2 The chromosome set of wheat somatic cells is 28. Determine the chromosome set and the number of DNA molecules in one of the ovule cells before meiosis, in meiosis anaphase I and meiosis anaphase II. Explain what processes take place during these periods and how they affect the change in the number of DNA and chromosomes.

Decision. Before us is a classic, well-known problem in cytology. It is important to remember here: if the task asks you to determine the chromosome set and the number of DNA molecules, besides, they show numbers - do not limit yourself to the formula: be sure to indicate the numbers.

The following steps are required for the solution:

1. indicate the initial number of DNA molecules. In this case, it is 56 - since they double, and the number of chromosomes does not change;


2. describe the anaphase of meiosis I: homologous chromosomes diverge to the poles;


3. describe the anaphase of meiosis II: the number of DNA molecules is 28, chromosomes - 28, sister chromatids-chromosomes diverge to the poles, since after the reduction division of meiosis I, the number of chromosomes and DNA decreased by 2 times.

In this formulation, the answer is likely to bring the desired high score.

Example 3 What chromosome set is typical for pine pollen grain and sperm cells? From what initial cells and as a result of what division are these cells formed?

Decision. The problem is formulated transparently, the answer is simple and easily broken down into components:

1. pollen grain and sperm cells have a haploid set of chromosomes;


2. pollen grain cells develop from haploid spores - by mitosis;


3. sperm - from pollen grain cells (generative cells), also by mitosis.

Example 4 Cattle have 60 chromosomes in their somatic cells. Determine the number of chromosomes and DNA molecules in ovarian cells in the interphase before the beginning of division and after meiosis I division. Explain how such a number of chromosomes and DNA molecules are formed.

Decision. The problem is solved according to the previously described algorithm. In the interphase before the start of division, the number of DNA molecules is 120, chromosomes - 60; after meiosis, I-respectively 60 and 30. It is important to note in the answer that before the start of division, the DNA molecules are doubled, and the number of chromosomes does not change; we are dealing with reduction division, so the number of DNA is reduced by 2 times.

Example 5 What chromosome set is typical for the cells of the outgrowth and gametes of the fern? Explain from what initial cells and as a result of what division these cells are formed.

Decision. This is the same type of problem where the answer is easily decomposed into three elements:

1. indicate the set of germ chromosomes n, gametes - n;


2. be sure to indicate that the outgrowth develops from a haploid spore by mitosis, and gametes - on a haploid outgrowth, by mitosis;


3. since the exact number of chromosomes is not indicated, you can limit yourself to the formula and write simply n.

Example 6 Chimpanzees have 48 chromosomes in somatic cells. Determine the chromosome set and the number of DNA molecules in cells before meiosis, in meiosis anaphase I and in meiosis prophase II. Explain the answer.

Decision. As you can see, in such tasks, the number of response criteria is clearly visible. In this case, they are: determine the set of chromosomes; define it in certain phases - and be sure to give explanations. It is most logical in the answer to give explanations after each numerical answer. For example:

1. we give the formula: before meiosis, the set of chromosomes and DNA is 2n4c; at the end of the interphase, DNA doubling occurred, the chromosomes became two-chromatid; 48 chromosomes and 96 DNA molecules;


2. in the anaphase of meiosis, the number of chromosomes does not change and is equal to 2n4c;


3. meiotic prophase II is entered by haploid cells having a set of two-chromatid chromosomes with a set of n2c. Therefore, at this stage we have 24 chromosomes and 48 DNA molecules.

A new study guide is offered to the attention of students and teachers, which will help them successfully prepare for the unified state exam in biology. The handbook contains all the theoretical material on the biology course required for passing the exam. It includes all elements of the content, checked by control and measuring materials, and helps to generalize and systematize knowledge and skills for the course of the secondary (complete) school. The theoretical material is presented in a concise, accessible form. Each section is accompanied by examples of test tasks that allow you to test your knowledge and the degree of preparedness for the certification exam. Practical tasks correspond to the USE format. At the end of the manual, answers to tests are given that will help schoolchildren and applicants to test themselves and fill in the gaps. The manual is addressed to schoolchildren, applicants and teachers.

You can learn anything, but it is more important to learn how to reflect and apply the learned knowledge. otherwise, you will not be able to score adequate passing scores. during the educational process, pay attention to the formation of biological thinking, teach students to use an adequate language for the subject, work with terminology. There is no point in using a term in a textbook paragraph if it does not work in the next two years.