Tasks 2 parts in biology. Preface. Training options for the exam in biology

Training options for the exam in biology

After the thematic tasks in biology, start practicing. Since in order to demonstrate a high level of knowledge, it is necessary to confidently work with diagrams, tables and graphs. Explain biological processes using graphical information.

First of all, download FIPI, which is a sample and gives an idea of the structure and form of complexity of future assignments for the exam.

Based on the new demo developed 10 training options, register and track the level of knowledge in personal account.

Identify, analyze mistakes and train again. Your success is the constant decision of options during preparation!

The USE test in biology 2019 includes 28 tasks.

Part 1 contains 21 tasks with a short answer (a sequence of numbers, a number, a word or a phrase)
Part 2 contains 7 tasks with a detailed answer (give a complete answer: explanation, description or justification; express and argue your own opinion).

The variant is thematically grouped.

The first part contains 21 tasks, which are grouped according to the content blocks presented in:
- For multiple choice;
- To establish correspondence;
- To establish the sequence of processes or phenomena;
- Tasks in cytology and genetics;
- To complement the drawings;
- Analysis of a schema or table.
The second part contains 7 tasks. For successful solution which the student is required to thoroughly master the conceptual apparatus and competently operate with biological terms.

A short analysis of the conditions of some tasks

Tasks from the block of the first ticket:

- a biological fragment is presented, requiring the establishment of links between concepts;
- count the number of chromosomes and establish the number of cells formed during various processes;
- find examples in the text that correspond to the concepts;
- to test knowledge of species properties - select criteria from the test that correspond to the species.

Assessment of tests in biology USE

Behind first part ticket maximum - 38 points.
For problem solving second part - 20 points.

The points received for correctly completed tasks are summed up.

Converting points to grades

0-35 points - 2,
36-54 points - 3,
55-71 points - 4,
72 and above points - 5;

For admission to a budget place in prestigious university you need to score more than 84 points.

Decide! Dare! Strive for the best!

We advise you to stop any attempts to solve any tasks from the USE samples in the evening on the eve of the exam - nothing but hassle will come of this. Relax, take a walk - in general, try to distract yourself and get a good night's sleep on the eve of the exam. Gather everything you need in advance: passport, black gel pens, replacement shoes (if required), etc.

Listen carefully to all instructions before starting the exam: statistics show that at least one technical Every second dealer makes a mistake.

Timing

	1 hour
	40 minutes
	30 minutes
	30 minutes
	20 minutes

Draft of part C

We advise you to start with part C (it’s better not to even open parts A and B at first), since it is the most difficult, and leaving it to the end, when you have already solved so many tasks on different topics and there is confusion in your head, is unreasonable.

When formulating answers, mentally count your statements: there must be at least two of them in task C1 and at least three in tasks C2-C6, or better, more - you don’t know which of your statements will match the inspector’s sample answer.

Sometimes it is useful to start by defining the biological concepts contained in the question itself - it may well turn out that you already get one point for knowing them.

Try to use scientific language rather than answer questions in everyday style. Consider that you are writing a paragraph from a biology textbook on this subject.

When solving genetic and cytological problems, it is not enough to write crossover schemes and sequences of nucleotides or amino acids - do not forget to accompany all these calculations with your detailed reasoning - then you will not find fault.

In tasks for finding and correcting errors, do not forget that there should always be three errors. When correcting errors, you cannot use simple denial, you need to make your own true statement, demonstrating your knowledge in this area.

Draft of parts A and B

As a draft of parts A and B, you can use the task sheets themselves, you can write anything there: checkmarks, numbers, letters, cross them out, make drawings, etc. - USE assignments with your notes are handed over, but not checked.

Now read the following sentence briefly:

BOYS IRGALI IN THE YARD IN PYANTASHKI

Did you notice anything? Try to find 3 misspellings of the words as you read more carefully.

This is a model of how our brain works. He always grasps the main thing, not paying due attention to the secondary. Test writers make extensive use of this feature of our thinking, hoping that we will be inattentive to the details of the formulated task.

In order not to fall for this bait, do not be lazy to read the tasks of parts A and B several times and make a choice of answers, carefully weighing all the pros and cons. The choice argument "I think so" or "because it's wrong" won't work. It is better to say all the possible arguments without turning the whole process of thinking into an unconscious choice.

Sometimes it is useful to make sketchy drawings or even write individual phrases.

If you do not know the answers to some questions, it is better to answer at random than to leave an empty box. In the second case, you will definitely get a "0" for this question, and in the first case, there is a chance to accidentally hit the right answer. For incorrect answers in parts A and B, points are not reduced.

Examples of USE assignments in biology

part 2 with explanations

Many tasks of part 2 require not only good knowledge biological material, but, unlike other tasks, the ability to clearly and clearly state the data, that is, good skills in Russian are required. This imposes additional difficulties in the answer. However, even if you know the answer to the question posed (or think you do), your arguments must match those in the inspector's cheat sheet. This is a huge minus of the USE: if you answer orally, then in the process of talking with the teacher, you can correct, correct your answer, give additional arguments, etc., this will not affect your grade, often your rating will even increase. When completing the USE assignments, what you write will not be changed, you will not add: “what is written with a pen cannot be cut down with an ax.”

Try to solve these examples yourself first,

then carefully study the explanations for them and compare the results.

What relationship is established between algae and fungus in the lichen thallus? Explain the role of both organisms in this relationship.

Explain why earthworms avoid waterlogged areas of soil and crawl out onto its surface.

What is the purpose of artificial mutagenesis?

What numbers indicate the vena cava in the figure? What number denotes the veins that carry arterial blood? What number indicates the vessel that receives blood from the left ventricle?

5. Find the errors in the given text, correct them, indicate the numbers of the sentences in which they are made, write down these sentences without errors.

1. All living organisms - animals, plants, fungi, bacteria, viruses - are made up of cells. 2. Any cells have a plasma membrane. 3. Outside the membrane, the cells of living organisms have a rigid cell wall. 4. All cells have a nucleus. 5. The cell nucleus contains the genetic material of the cell - DNA molecules.

6. Describe the features of the plant kingdom. Give at least 3 signs.

What physiological changes can occur in a person who works all his life on a lathe? Give at least three examples.

Why is it necessary to constantly synthesize in the cells of the human bodyorganic matter?

Establish a sequence of processes characteristic of leaf fall.

1) the formation of a separating layer on the petiole

2) accumulation of harmful substances in the leaves during the summer

3) leaf fall

4) destruction of chlorophyll due to cooling and a decrease in the amount of light

5) changing the color of the leaves.

10. Establish the correct sequence of stages of development of the liver

fluke starting from the zygote.

1) cyst

2) egg

3) ciliary larva

4) tailed larva

5) zygote

6) an adult worm.

11. Describe the stages of natural selection that leads to the preservation of individuals with an average value of a trait.

12. Prove that the cell is an open system.

13. What chromosome set is typical for the nuclei of epidermal cells

leaf and eight-core embryo sac of the ovule of a flowering plant?

Explain from what initial cells and as a result of what division these cells are formed.

cells.

14. It is known that all types of RNA are synthesized on a DNA template. The fragment of the DNA molecule, on which the region of the central loop of tRNA is synthesized, has the following nucleotide sequence ATAGCTGAACGGACT. Set the nucleotide sequence of the t-RNA section that is synthesized on this fragment and the amino acid that this t-RNA will transfer during protein biosynthesis, if the third triplet corresponds to the t-RNA anticodon. Explain the answer. To solve the problem, use the table of the genetic code.

First

base

Second base

Third

base

hair dryer

Lei

Ser

Tyr

cis

Three

Lei

Pro

gis

Gln

Arg

ile

Met

Tre

Asn

Liz

Ser

Arg

Shaft

Ala

Asp

Glu

gli

15. Drosophila somatic cells contain 8 chromosomes. Determine how many chromosomes and DNA molecules are contained in the nuclei during gametogenesis before division in the interphase and at the end of the telophase of meiosis I. Explain how such a number of chromosomes and DNA molecules are formed.

16. In humans, the gene that causes one of the forms of hereditary deaf-mutism is recessive with respect to the gene for normal hearing. From the marriage of a deaf-mute woman with a normal man, a deaf-mute child was born. Determine the genotypes of all family members.

17. In humans, albinism and the ability to predominantly use the left hand are recessive traits that are inherited independently. What are the genotypes of parents with normal pigmentation and who own right hand if they have an albino and left-handed child?

Explanations for the tasks of part 2

1. Answer to this question(No. 22 in the test) should normally contain two arguments (sentences), although three sentences are possible, and even more if the arguments are correct; in the end you can get 2 primary scores. This question is an example of the so-called. "specific" question: the answers to it are unambiguous and clear. Here are two possible answers:

Algae is an autotrophic organism, it synthesizes organic substances, part of them gives to the fungus. 2. A fungus is a heterotrophic organism, its hyphae absorb water with mineral salts, which the fungus shares with algae. 3. So, the lichen is a symbiotic organism.

Or: 1. A symbiotic relationship is established between an alga and a fungus (a lichen is a symbiont). 2. Algae gives the fungus a part of what it synthesizes during photosynthesis organic matter, the fungus supplies the algae with water and minerals.

Important: state your arguments point by point, not in a continuous text. Note that in either case, the answer contains two necessary key components: a symbiotic relationship between organisms and an explanation of what the meaning of this symbiotic relationship is. Without these components, the answer will not be correct!

2. There are two keywords in the question: "why" and "waterlogged", if you explain them correctly, you will get two points. So: 1. Waterlogged areas of soil contain little oxygen. 2. Since earthworms breathe through moist skin, they crawl to the surface to breathe atmospheric air.

Or: 1. A waterlogged kidney contains little oxygen. 2. Under such conditions, the worms cannot supply themselves with sufficient oxygen, as they breathe through the skin. 3. Therefore, they crawl to the surface to absorb oxygen from the atmosphere.

Important: always clearly analyze the question, highlight keywords, which will form the basis of your answer. In other words, answer only the question posed, there is no need to state details that are not relevant, this will divert your time and energy, and will not add points.

An example of a "non-specific" question (not the most "terrible"), and this is its complexity: what and how to answer? Key words: “for what purpose” and “artificial mutagenesis (IM)”, we move on from them. In this question, it is best to start by clarifying the concept of "artificial mutagenesis", since the rest of the answer comes from it.

1. MI is the process of getting a large number mutations with the help of enhancer factors (mutagenic factors). 2. Thanks to MI, there is a significant increase in the number and diversity of genotypes and phenotypes of individuals, among which individuals with traits useful for humans can be selected.

Or: 1. MI is… (see above). 2. IM is actively used in agricultural practice, as it allows obtaining organisms with new and useful traits for humans. 3. Thanks to IM, it is possible to increase the resistance and endurance of plants, to obtain high-yielding polyploid forms, etc.

Important: often graduates begin to deviate from the question posed, for example, they begin to talk about the types of artificial mutagenesis, to cite specific achievements. This does not apply to the content of the assignment! Or they do not give a definition of MI, then they get one point that “hangs in the air”; painful thoughts begin: what else to write? As a result, time is lost, but one point is obtained.

the task is of the “specific” type, to complete it you need to know accurate and clear knowledge of the cardiovascular system, so those who know the material well will win. The answers are unambiguous: 1. The vena cava are numbered 2 and 3 (2 - superior vena cava, 3 - inferior). 2. Veins with arterial blood - 5, these are pulmonary veins. 3. Blood from the left ventricle enters the aorta - 5).

Correct answers: Wrong sentences No. 1,3,4.

1(1). Almost All living organisms - animals, plants, fungi, bacteria - are made up of cells. Or: Living organisms - animals, plants, fungi, bacteria - consist of cells, except for viruses. 3(2). Outside the membrane, plant, fungal, and bacterial cells have a rigid cell wall. Or: Outside the membrane, the cells of all living organisms, except animals, have a rigid cell wall. 4(3). All cells except bacteria have a nucleus. Or: Eukaryotic cells (plants, animals, and fungi) have a nucleus.

The task also refers to specific ones, since there are clear and precise criteria (signs) of each kingdom. But there is also a difficulty: there are many signs of plants, and the question is three-point, that is, three points (sentences) must be indicated. In a similar question, you should do this: you should indicate not three, but more answers, even five or six. But it is necessary to take into account which features are more important, which ones are less important, you need to start with the more important ones: 1. All plants are autotrophs, they produce organic substances through photosynthesis. 2. The process of photosynthesis takes place with the participation of special pigments, in most plants this pigment is chlorophyll. 3. Pigments are located in two-membrane organelles - chloroplasts or in chromatophores. 4. A plant cell has a cell wall made of carbohydrates and large vacuoles. 5. The main reserve substance of plants is carbohydrate starch. 6. Most plants are not capable of active movement. 7. Plants have unlimited growth. 8. Plants have a complex life cycle, consisting of alternating sporophyte and gametophyte generations.

The most important signs, without which it is impossible to get the highest score, are: 1, 2, 3, 4, but if you indicate other signs, listing them separated by commas, in one sentence, it will be fine.

This question is creative. It is unlikely that you will find an answer to it in any textbook for preparing, which means that you need to think carefully about what to answer. It is necessary to evaluate what the work of a turner is and draw conclusions from this.

1. It is possible to develop varicose veins due to constant standing on your feet and stagnation of blood in the veins. 2. Visual impairment is possible, as it is necessary to constantly control the dimensions of parts with a lack of lighting. 3. Constant noise from working machinery can lead to hearing loss. 4. Since you often have to bend over the part being made, damage to the musculoskeletal system, especially the spine, is likely.

Important: in such a task, there is a high probability of an incorrect answer due to vagueness, for example, a student writes: “the condition of the vessels will worsen” - which ones? why? - unclear; or “there is a deposition of harmful substances in the body” - what substances, why? Where are they deposited? – again unclear. Remember: you must answer very clearly and specifically, not “smear”, avoid vagueness and uncertainty. Schoolchildren often write in this way, especially if they do not know the exact answer.

Another example of a "non-standard" question; on the one hand, it is based on data that are presented in various topics of the general biology course, on the other hand, there is no specific answer to this question. The key words are “why” and “constantly synthesize organic substances”. We will proceed from them. First, let's think about what substances the cells must constantly synthesize, these are ATP, proteins, reserve lipids and carbohydrates, hormones. Why should they be synthesized? Yes, because they are constantly consumed or excreted from the body.

Having made such considerations, we can proceed to the answer: 1. A constant synthesis of ATP is needed, which is consumed in the processes of plastic exchange. 2. When energy exchange there is a breakdown of organic substances, especially carbohydrates and lipids, they must be obtained from food. 3. Part of the substances leaves the body in the process of excretion, which means that their continuous replenishment is required.

An example of a fairly simple question, even without knowing the material, you can logically reach the correct answer. The sequence of stages is as follows: 1. accumulation of harmful substances in the leaves during the summer. 2.destruction of chlorophyll due to cooling and a decrease in the amount of light. 3. discoloration of the leaves. 4. formation of a separating layer on the petiole. 5. leaf fall.

This question is more complex than the previous one, but it is not difficult in the sense that it requires only one thing - a good knowledge of the life cycle of the liver fluke, nothing more. Answer: 5) zygote. 2) an egg. 3) ciliary larva. 4) tailed larva. 1) cyst. 6) an adult worm.

Another creative question. It is based on the relatively small topic of natural selection, but in this case it is very important how this issue is correctly, clearly and clearly stated. Let's briefly consider the logic of the answer. Let's highlight the key words: "stages of natural selection" and "average value of a trait." First, let's recall the EO mechanism. Since our sign preserves the mean value, it means that we need to describe the stabilizing form of the EO. Where does the process begin? From what Darwin wrote about: from the appearance of random hereditary changes (mutations). These mutations concern a variety of traits and the magnitude of their changes is also different; it is important that these changes appear completely by chance. Natural selection acts on a variety of new phenotypic traits. As a result, individuals with imperfect, unsuitable traits will die, while individuals whose traits meet these conditions will remain. Recall that the selection in our problem is stabilizing, that is, it operates under relatively constant (stable) conditions. With this form of selection, individuals with the average value of the trait will remain, as well as possible well adapted to these conditions ...

These should be your reasoning, now we will state them in the answer: 1. As a result of hereditary changes (mutations), individuals with the most different meaning sign. 2. The signs are affected by natural selection, as a result of which individuals with the most appropriate signs for the given conditions remain. In this case, there is a stabilizing form of selection that operates under constant environmental conditions. 4. As a result, individuals with an average trait remain, as the most adapted to given conditions.

Or: 1. Natural selection retains those features that provide the body with the best adaptation to the environment. 2. New characters randomly appear as a result of hereditary variability. 3. Signs with an average value are preserved with a stabilizing form of EO under constant conditions.

In any form of answer, we used keywords, without which the answer would not be complete. We have carefully analyzed the material on which the answer is based. With similar answers, in our opinion, the first option is more preferable, since it presents the material more consistently, in the order of evolutionary events.

The complexity of this task is that not everyone knows what an “open system” is, not all textbooks and teachers pay attention to this concept, but this is a key expression. Therefore, it is better to start the answer with the definition of an open system, only then the answer will be complete: 1. An open system is a system that is in constant exchange of compounds and energy with the environment. 2. The cell receives from environment the substances it needs (proteins, fats, carbohydrates, etc.) and releases metabolic products. 3. During the decay of substances in the cell, energy is released and stored, part of the energy is dissipated in the environment in the form of heat. (As an additional point, one can point to the information exchange: 4. From the external environment, the cell receives certain information, reacts to it, and delivers its information to the environment through substances and other signals.).

A task that confuses many students. Why, after all, there is nothing particularly complicated in it? Firstly, the course of botany is studied in the 6th grade and when preparing for it, “hands do not reach”. Secondly, a number of students have a not very serious attitude to the subject (“pistils-stamens”). Thirdly, this question is a combination of botany and general biology (mitosis, meiosis), that is, a complex, general biological question. To answer it well, you need to know the following: a. the concept of cell ploidy; b. the process of mitosis; in. the process of meiosis; d. life cycle of angiosperms; e. the mechanism of development of the ovule in the ovary.

The cells of the leaf epidermis are diploid (2 n ), the cells of the eight-nuclear embryo sac of the ovule are haploid ( n ). 2. Epidermal cells are obtained as a result of mitosis from a diploid zygote. 3. When the embryo sac is formed, meiosis of the initial diploid cell first occurs, and then the embryo sac is obtained from the resulting haploid microspore cell as a result of three successive mitoses. Or:

The cells of the leaf epidermis are characterized by a diploid set of chromosomes (2 n ), since all vegetative cells of a plant are formed as a result of multiple mitosis from a diploid zygote. 2. An eight-nuclear embryo sac begins to form in the ovule of the ovary from a diploid cell (2 n ), which is divided by meiosis.3. 4 microspore cells are obtained ( n ), from one as a result of three successive divisions of mitosis, which means that 8 haploid cells are formed.

Conclusion: when preparing, there are no unimportant or unloved topics. All without exception (!) Topics are important, and you need to know them.

This type of task causes significant difficulties, although the material on which the task is based is small: the topic "Cellular biosynthesis: replication, transcription, translation." But not only is it important know material, but be able to apply for solving practical problems.

The key proposition that must be taken into account when deciding is "all types of RNA are synthesized on a DNA template." This means that the tRNA fragment will be synthesized in the original DNA region: ATAGCTGAACGGACT. The desired t-RNA sequence is determined according to the principle of complementarity, taking into account that instead of T, the t-RNA will include Y: UAUCGATSUUGCCUGA. Next, on the found t-RNA sequence, we find the third triplet: CUU is an anticodon. In the process of protein synthesis, according to the principle of complementarity, it will interact with the corresponding mRNA triplet. The desired mRNA triplet is GAA. According to the table of the genetic course, we find which amino acid this triplet encodes: Glu (glutamine).

Basic mistake when solving problems of this type: students first “find” i-RNA based on the original DNA, as they are used to doing this when solving other tasks, and only then - t-RNA, etc. The solution turns out to be incorrect (the key sentence is not taken into account! ).

This type of problem is also quite difficult to solve, although the material used is small: the topics "Mitosis" and "Meiosis", the topic "Ploidy of cells" is also useful. Solution: the original somatic cells of Drosophila contain 8 chromosomes (4 homologous pairs). Very important: the chromosomes of these cells are single, so the cells themselves are diploid (2 n ), because they contain 4 pairs of homologues, and the number of DNA molecules is 8.

This genetic task is not among the complex ones. Crossing is monohybrid, since one sign is considered - deaf-mutism. Since dominant and recessive alleles are directly indicated, the decision is even more simplified. Let's denote the alleles: BUT - normal hearing a - deaf-mute. According to the condition of the problem, the woman is deaf and dumb, so her genotype is aa and it cannot be otherwise, a man has a genotype Ah . (Why not AA “The fact is that with the AA genotype, the genotypes of children will be only Aa and they will have hearing, and, according to the condition of the problem, the child is sick). Now let's make a crossover scheme:

R: aa*aa

G : a aa

F 1 :Aa aa - deaf child.

17. This task is more complicated, two signs are considered, that is, a dihybrid cross. To solve the problem, you can use the following rule: consider each trait separately from the other, otherwise it is difficult to select the genotypes of the parents.

I. Consider the sign of pigmentation:

BUT – normal pigmentation, a - albinism - according to the condition of the problem.

Since the child is an albino (recessive trait), therefore, his genotype for this trait is aa . Each parent has normal pigmentation, which means they both carry the dominant allele. BUT .But since they have a child with the genotype aa , then each of them must also carry the recessive allele a . Therefore, the genotype of the parents for the pigmentation gene is Ah .

II. The second sign is possession of the right-left hand:

AT - right handedness b - left-handedness - according to the condition of the task.

The child is left-handed (recessive trait), therefore, its genotype is bb .Parents are right-handed, which means that each of them carries a dominant allele AT . Their child is left handed bb ), so each parent carries the recessive allele b . Therefore, the genotypes of the parents for this trait are Вb . Therefore, the mother's genotype is AaBb ; father's genotype AaBb ; child's genotype aabb .

The scheme of marriage is as follows:

Thus, the parents are heterozygous for each pair of traits and their genotype is AaBb .

( Note: there are always problems in genetics, they are quite diverse, so their analysis should be given special and considerable attention. I or my colleagues will be happy to help you with this issue).

USE questions,

most challenging for graduates

(Note: The ten questions below are explained by the ideal answers used by the members of the review committee for assessment. There are many complexities and hidden "pitfalls" that require special and careful analysis. Otherwise, it often turns out that the student, in his opinion, did everything perfectly, because he knows the material and presented it well (in his opinion!) And in the end, instead of the coveted 3 points, he gets 1 .... For help, contact us, specialists).

one). Task 22: What harm does smoking do to the respiratory and cardiovascular systems? Explain the answer.

2). Task 22 : Is it possible to breed chickens from eggs bought at the grocery store that came from a poultry farm? Explain the answer.

3). Task 22: Why do thick seedlings of carrots and beets need to be thinned out to get a good harvest? Explain the answer.

4). Task 25: What features of angiosperms contribute to their prosperity on Earth? List at least four features and explain your answer.

5). Task 25: Whales constantly live in the water, have a streamlined body shape and other adaptations for life in this environment. What signs of these animals indicate that they are secondary aquatic organisms? List at least four features.

6). Task 26: Why, when pesticides are used to control agricultural pests, besides the pests themselves, other animals die, but more often predators, rather than herbivores?

Give three reasons.

7). Task 26: Why does not even a long-term effect of stabilizing selection on individuals of the same species lead to the formation of complete phenotypic uniformity? Justify your answer with three arguments.

eight). Task 27: What set of chromosomes does the embryo and endosperm of a diploid plant have? From what initial cells and as a result of what division did they form?

nine). Task 27: What chromosome set is typical for fern leaf and spore cells? Explain from what initial cell and as a result of what division a set of chromosomes is formed in each cell.

ten). Task 27: Drosophila somatic cells contain 8 chromosomes. Determine how many chromosomes and DNA molecules are contained in the nucleus during gametogenesis before the start of division and in metaphaseImeiosis. Explain your results. Describe the behavior of chromosomes in metaphaseImeiosis.

one). Task 22: Answer elements:

The particles of tobacco smoke and tar deposited on the walls of the respiratory tract and lungs reduce the efficiency of gas exchange and the supply of oxygen to the blood.

Penetrating into the blood toxic substances (nicotine) narrow the lumen of blood vessels, increase pressure, thus increasing the load on the heart.

2). Task 22: Answer elements:

It is forbidden

The store receives from the poultry farm unfertilized eggs of hens, in which there are no embryos, or chilled eggs. in which the fetus died.

3). Task 22: Answer elements:

These plants form root crops, the formation of which requires a large volume of soil.

Thinning plants reduces competition, promotes root development and leads to higher yields.

4). Task 25: Answer elements:

The presence and diversity of flowers (inflorescences), ensuring their adaptability to pollination by wind, animals, water, self-pollination.

Presence and variety of fruits that protect the seeds and promote their spread.

Well-developed conducting system (vessels and sieve tubes) that ensure the movement of substances throughout the plant

The presence of various modifications (shoots, roots) that ensure vegetative propagation and adaptability to different environmental conditions.

A variety of life forms (trees, grasses, shrubs) that provide adaptations to different conditions habitat.

The presence of a triploid endosperm ensures the germination of the seed, and the embryo with a large amount of nutrients.

5). Task 25: Answer elements:

They breathe atmospheric air, have lungs.

The presence of a belt and terrestrial-type limbs modified into fins.

Presence of rudiments of the hind limb girdle.

The hairline is reduced.

6). Task 26: Answer elements:

Pesticides do not have a selective effect and affect not only pests, but also other animals, causing their death.

The biomass of the lower trophic levels is much higher than the upper level, so the concentration of pesticides in the bodies of herbivores is lower than in the bodies of predators.

Predators are at the highest trophic level in food chains, they eat herbivores, which leads to the concentration of pesticides in their bodies and, as a result, to the death of predators.

7). Task 26: Answer elements:

There is a wide norm of modification (phenotypic) variability, which leads to a variety of phenotypes.

Combinative variability (sexual process and gene recombination) leads to genotypic and phenotypic diversity, to the manifestation of recessive traits.

In populations, new mutations constantly arise, which accumulate and change the phenotypic composition of the population.

Genetic drift and population waves change the frequency of alleles. A set of genes and traits.

Migrations of individuals change the genotypic and phenotypic composition of the population.

eight). Task 27: Answer elements:

Chromosomal set in embryo-2n, in the endosperm - 3n.

The embryo is formed from a zygote that divides by mitosis.

The endosperm is formed from the fertilized central cell by mitotic division.

nine). Task 27: Answer elements:

In fern leaf cells, a diploid set of chromosomes-2n

The fern leaf develops from the zygote (sporophyte) by mitosis.

In a fern spore, the haploid set of chromosomes isn.

Spores are formed from sporangium cells as a result of reduction division (meiosis).

ten). Task 27: The scheme for solving the problem includes:

Before the start of division, the number of chromosomes is 8, the amount of DNA is 16.

Before the start of division, the DNA molecules double, each chromosome consists of two sister chromatids, but the number of chromosomes does not change.

In metaphaseImeiosis, the number of chromosomes is 8, DNA molecules - 16.

In metaphaseImeiosis, the number of chromosomes and DNA does not change, pairs of homologous chromosomes are located in the equatorial zone above and below the plane of the equator of the cell.

Master Class
"Preparation for the OGE in Biology"
Analysis of tasks No. 28; No. 31; No. 32 of the second part of the OGE in biology

The master class was developed and conducted by a biology teacher of the 1st category

Isakova Natalya Vladislavovna

(MOU "Emmausskaya secondary school", Tver)

Goals and objectives of the master class in biology

1. Study the specification of control and measuring materials in biology in the OGE 2016-2017.

2. Familiarize yourself with the content elements codifier and the requirements for the level of graduate training educational institutions for OGE in biology.

3. To form the skills and abilities of working with tasks basic level difficulties.

Equipment: presentation, demo OGE in biology 2016-2017; codifier of content elements and requirements for the level of training of graduates of general educational institutions for the OGE 2016-2017 in biology (draft); CMM specification (draft).

Introductory speech of the teacher

Let's get acquainted with the instructions for performing the work:

The examination paper consists of two parts, including 32 tasks. Part I contains 28 tasks with a short answer, part II contains 4 tasks with a detailed answer.

For execution examination work 3 hours (180 minutes) are allotted.

Answers to tasks 1-22 are written as one digit, which corresponds to the number of the correct answer. Write this figure in the answer field in the text of the work, then transfer it to the answer sheet No. 1.

Answers to tasks 23-28 are written as a sequence of numbers. Write down this sequence of numbers in the answer field in the text of the work, then transfer it to the answer sheet No. 1.

For tasks 29-32, a detailed answer should be given. Tasks are performed on the answer sheet No. 2.

When completing assignments, you can use a draft. Draft entries do not count towards the assessment of the work.

The points you get for completed tasks are summed up. Try to complete as many tasks as possible and score the largest number points.

Let's get acquainted with the criteria for assessing the examination work.

1. For the correct performance of tasks 1-22, 1 point is given.

2. For the correct answer to each of the tasks 23-27, 2 points are given.

3. For the answer to tasks 23-24, 1 point is given if the answer contains any two digits presented in the standard answer and 0 points if only one digit is correctly indicated or none is indicated.

4. If the examinee indicates in the answer more characters than necessary, then 1 point is reduced for each extra character (to 0 points).

5. For a complete correct answer to task 28, 3 points are given; 2 points are given if any position of the answer contains a different character than that presented in the standard answer; 1 point is given if any two positions of the answer contain characters other than those presented in the standard answer, and 0 points in all other cases.

So, let's proceed to the analysis of task No. 28 of the second part of the OGE in biology.

In order for our work to be productive, let's remember leaf shapes, leaf types (venation).

(slides 3-11).
We begin the analysis of task No. 28 of the standard version No. 1 of 2017.

1. Determine the leaf type If the leaf has a petiole, then the leaf type is petiolate. If the leaf has no petiole, the leaf is sessile.

2. We determine the venation of the leaf. This leaf has pinnate (netted) venation, since the leaf has one powerful vein located in the middle.

3.Sheet shape. In order to determine the shape of the sheet, it is necessary to draw dotted lines in KIME in the figure and, using the sample presented in the task, determine the shape of the sheet (working with an interactive whiteboard).

4. Determine the type of sheet using a ruler and a pencil. Draw dotted lines. If the length is equal to or exceeds the width by 1-2 times, then the type of sheet in terms of the ratio of length and width is ovoid, oval or obovate. This sheet has a width of 3 cm, a length of 4.5 cm, which means the shape of this sheet is ovoid board).

5. It remains to determine the edge of the sheet. We will carefully consider the samples proposed in KIME, we will choose a suitable sample for our sheet. The sheet is entire-marginal, as there are no teeth along the edges.

Task #28 completed successfully.

We proceed to the analysis of task No. 31 of the typical option No. 1 of 2017.

1. Carefully read the condition of the problem.

2. We emphasize in the text the questions that we must answer (what we must calculate).

3. So we need to indicate: the energy consumption of the morning workout, the recommended dishes, the calorie content of lunch and the amount of protein in it

4. Let's calculate the energy consumption of the morning workout. To do this, we will use table No. 3 "Energy costs for various types of physical activity."

From the condition of the problem, we know that Olga plays tennis, which means that the energy cost will be 7.5 kcal/min.

(the value is taken from table No. 3, taking into account the sport that Olga is involved in). We need to calculate the energy consumption of a two-hour morning workout. So 120 minutes (2 hours of training) X 7.5 kcal. (energy cost of physical activity) and get the energy consumption of a morning workout of 900 kcal. (120 X 7.5 \u003d 900 kcal.)

To compile the menu, we use table No. 2 of the column "Energy value".

Menu: Meatball sandwich (energy value - 425 kcal.)

Vegetable salad (energy value-60 kcal.)

Ice cream with chocolate filling (energy value - 325 kcal.)

Tea with sugar (two teaspoons) - energy. value - 68 kcal.

+60 + 325 + 68 = 878 kcal. (calorie content of the recommended lunch).

6. Find the amount of protein in the recommended lunch. To do this, we use table No. 2 of the column "Squirrels".

39+ 3 +6= 48

Job completed successfully.
Let's analyze task No. 32 of the typical option No. 1 of 2017.
Why did the coach pay special attention to Olga's protein content in the ordered dishes? Specify at least two arguments.

Task #32 follows from task #31.
1. Protein is the main building material for the body. Protein is made up of muscles, ligaments, skin and internal organs.

2. Protein is a source of energy.
Job done, thanks for the job!

BIOLOGY

Under the editorship of Academician of the Russian Academy of Medical Sciences, Professor V.N. Yarygin

In two books

Book 2

Fifth edition, corrected and supplemented Recommended by the Ministry of Education of the Russian Federation

as a textbook for students of medical specialties of higher educational institutions

Moscow " graduate School» 2003

UDC 574/578 BBK 28.0 B63

V.N. Yarygin, V.I. Vasilyeva, I.N. Volkov, V.V. Sinelshchikova

Reviewers:

Department of Medical Biology and Genetics of the Tver State Medical Academy (Head of the Department - Prof. G. V. Khomullo);

Department of Biology of the Izhevsk State Medical Academy (Head of the Department

Prof. V.A. Glumov)

ISBN 5-06-004590-0

The original layout of this publication is the property of the Vysshaya Shkola publishing house, and its reproduction (reproduction) in any way without the consent of the publisher is prohibited.

FOREWORD

This book is a continuation of the textbook "Biology" for students of medical specialties. It includes sections on biological patterns, which manifest themselves at the population-species and biogeocenotic levels of the organization of life on Earth.

The numbering of the chapters continues with the 1st book.

SECTION IV POPULATION AND SPECIES LEVEL OF LIFE ORGANIZATION

The previously considered biological phenomena and mechanisms related to the molecular-genetic, cellular and ontogenetic levels of life organization were spatially limited to a single organism (multicellular or unicellular, prokaryotic or eukaryotic), and temporally - to its ontogenesis, or life cycle. The population-species level of organization belongs to the category of superorganisms.

Life is represented by separate species, which are collections of organisms that have properties heredity and variation.

These properties become the basis of the evolutionary process. The mechanisms that determine this result are selective survival and selective reproduction of individuals belonging to the same species. AT natural conditions especially intensive reproduction occurs in populations that are the minimum self-reproducing groups of individuals within a species.

Each "of the once existing or living species is the result of a certain cycle of evolutionary transformations at the population-species level, initially fixed in its gene pool. The latter is distinguished by two important qualities. Firstly, it contains biological information on how this species can survive and leave offspring under certain environmental conditions, and secondly, it has the ability to partially change the content of biological information contained in it.

the basis of the evolutionary and ecological plasticity of the species, i.e. the ability to adapt to existence in other conditions that change in historical time or from territory to territory. The population structure of a species, which leads to the disintegration of the gene pool of the species into gene pools of populations, contributes to the manifestation in the historical fate of the species, depending on the circumstances, of both noted qualities of the gene pool - conservatism and plasticity.

Thus, the general biological significance of the population-species level consists in the implementation of the elementary mechanisms of the evolutionary process that determine speciation.

The significance of what is happening at the population-species level for public health is determined by the presence of hereditary diseases, diseases with an obvious hereditary predisposition, as well as pronounced features of the gene pools of different human populations. The processes taking place at this level, combined with the ecological features of various territories, form the basis of a promising area of modern medicine - the epidemiology of non-communicable diseases.

CHAPTER 10 BIOLOGICAL SPECIES. POPULATION STRUCTURE OF THE SPECIES

10.1. THE CONCEPT OF THE VIEW

A species is a set of individuals that are similar in basic morphological and functional characteristics, karyotype, behavioral reactions, have a common origin, inhabit a certain territory (range), in natural conditions interbreed exclusively with each other and at the same time produce fertile offspring.

The species affiliation of an individual is determined by its compliance with the listed criteria: morphological, physiological and biochemical, cytogenetic, ethological, ecological, etc. The most important features of a species are its genetic ( reproductive) isolation, which consists in the non-crossing of individuals of a given species with representatives of other species, as well as

genetic stability in natural conditions, leading to an independent evolutionary destiny.

Since the time of K. Linnaeus, the species has been the main unit of taxonomy. The special position of the species among other systematic units (taxa) is due to the fact that this is the group in which individual individuals exist really. As part of a species in natural conditions, an individual is born, reaches puberty and performs its main biological function: participating in reproduction, it ensures the continuation of the genus. Unlike species, taxa of superspecific rank, such as genus, order, family, class, phylum, are not real life organisms. Separating them into natural system organic world reflects