1. What methods of division are characteristic of eukaryotic cells? For prokaryotic cells?

Mitosis, amitosis, simple binary fission, meiosis.

Eukaryotic cells are characterized by the following division methods: mitosis, amitosis, meiosis.

Prokaryotic cells are characterized by simple binary fission.

2. What is simple binary fission?

Simple binary fission is characteristic only of prokaryotic cells. Bacterial cells contain one chromosome, a circular DNA molecule. Before cell division, replication occurs and two identical DNA molecules are formed, each of them attached to the cytoplasmic membrane. During division, the plasmalemma grows between two DNA molecules in such a way that it ultimately divides the cell in two. Each resulting cell contains one identical DNA molecule.

3. What is mitosis? Describe the phases of mitosis.

Mitosis is the main method of division of eukaryotic cells, as a result of which two daughter cells with the same set of chromosomes are formed from one mother cell. For convenience, mitosis is divided into four phases:

● Prophase. In the cell, the volume of the nucleus increases, chromatin begins to spiral, resulting in the formation of chromosomes. Each chromosome consists of two sister chromatids connected at the centromere (in a diploid cell - set 2n4c). The nucleoli dissolve and the nuclear membrane disintegrates. Chromosomes end up in the hyaloplasm and are arranged randomly (chaotically) in it. Centrioles diverge in pairs to the cell poles, where they initiate the formation of spindle microtubules. Some of the spindle threads go from pole to pole, other threads are attached to the centromeres of chromosomes and contribute to their movement to the equatorial plane of the cell. Most plant cells lack centrioles. In this case, the centers for the formation of spindle microtubules are special structures consisting of small vacuoles.

● Metaphase. The formation of the fission spindle is completed. Chromosomes reach maximum spiralization and are arranged in an orderly manner in the equatorial plane of the cell. A so-called metaphase plate is formed, consisting of two-chromatid chromosomes.

● Anaphase. The spindle strands shorten, causing the sister chromatids of each chromosome to separate from each other and stretch toward opposite poles of the cell. From this moment on, the separated chromatids are called daughter chromosomes. The cell poles have the same genetic material (each pole has 2n2c).

● Telophase. Daughter chromosomes despiral (unwind) at the cell poles to form chromatin. Nuclear shells form around the nuclear material of each pole. Nucleoli appear in the two formed nuclei. The spindle filaments are destroyed. At this point, nuclear division ends and the cell begins to divide into two. In animal cells, a ring constriction appears in the equatorial plane, which deepens until the separation of two daughter cells occurs. Plant cells cannot divide by constriction, because have a rigid cell wall. In the equatorial plane of the plant cell, the so-called median plate is formed from the contents of the vesicles of the Golgi complex, which separates the two daughter cells.

4. How do daughter cells receive identical hereditary information as a result of mitosis? What is the biological significance of mitosis?

In metaphase, bichromatid chromosomes are located in the equatorial plane of the cell. The DNA molecules in sister chromatids are identical to each other, because formed as a result of replication of the original maternal DNA molecule (this occurred in the S-period of interphase preceding mitosis).

In anaphase, with the help of spindle threads, the sister chromatids of each chromosome are separated from each other and stretched to opposite poles of the cell. Thus, the two poles of the cell have the same genetic material (2n2c at each pole), which, upon completion of mitosis, becomes the genetic material of the two daughter cells.

The biological significance of mitosis is that it ensures the transmission of hereditary characteristics and properties over a series of cell generations. This is necessary for the normal development of a multicellular organism. Due to the precise and uniform distribution of chromosomes during mitosis, all cells in the body are genetically identical. Mitosis determines the growth and development of organisms, restoration of damaged tissues and organs (regeneration). Mitotic cell division underlies asexual reproduction in many organisms.

5. Number of chromosomes - n, chromatids - c. What will be the ratio of n and c for human somatic cells in the following periods of interphase and mitosis. Match:

1) In the G 1 period, each chromosome consists of one chromatid, i.e. somatic cells contain a set of 2n2c, which for humans is 46 chromosomes, 46 chromatids.

2) In the G 2 period, each chromosome consists of two chromatids, i.e. somatic cells contain a set of 2n4c (46 chromosomes, 92 chromatids).

3) In prophase of mitosis, the set of chromosomes and chromatids is 2n4c, (46 chromosomes, 92 chromatids).

4) In the metaphase of mitosis, the set of chromosomes and chromatids is 2n4c (46 chromosomes, 92 chromatids).

5) At the end of anaphase of mitosis, due to the separation of sister chromatids from each other and their divergence to opposite poles of the cell, each pole has a set of 2n2c (46 chromosomes, 46 chromatids).

6) At the end of the telophase of mitosis, two daughter cells are formed, each containing a set of 2n2c (46 chromosomes, 46 chromatids).

Answer: 1 - B, 2 - G, 3 - G, 4 - G, 5 - V, 6 - V.

6. How does amitosis differ from mitosis? Why do you think amitosis is called direct cell division, and mitosis is called indirect?

In contrast to mitosis, amitosis:

● The nucleus divides by constriction without chromatin spiralization and spindle formation; all four phases characteristic of mitosis are absent.

● Hereditary material is distributed unevenly and randomly among daughter nuclei.

● Often only nuclear division is observed without further division of the cell into two daughter cells. In this case, binucleate and even multinucleate cells appear.

● Less energy is wasted.

Mitosis is called indirect division, because. Compared to amitosis, it is a rather complex and precise process, consisting of four phases and requiring preliminary preparation (replication, doubling of centrioles, energy storage, synthesis of special proteins, etc.). During direct (i.e. simple, primitive) division - amitosis, the cell nucleus, without any special preparation, is quickly divided by a constriction, and the hereditary material is randomly distributed between the daughter nuclei.

7. In the nucleus of a non-dividing cell, hereditary material (DNA) is in the form of an amorphous dispersed substance - chromatin. Before division, chromatin spirals and forms compact structures - chromosomes, and after division it returns to its original state. Why do cells make such complex modifications of their hereditary material?

DNA in the composition of amorphous and dispersed chromatin during division would be impossible to accurately and evenly distribute between daughter cells (this is exactly the picture that is observed during amitosis - the hereditary material is distributed unevenly, randomly).

On the other hand, if cellular DNA were always in a compacted state (i.e., as part of spiralized chromosomes), it would be impossible to read all the necessary information from it.

Therefore, at the beginning of division, the cell transfers DNA to the most compact state, and after division is completed, it returns it to its original state, convenient for reading.

8*. It has been established that in diurnal animals the maximum mitotic activity of cells is observed in the evening, and the minimum - during the day. In animals that are nocturnal, cells divide most intensively in the morning, while mitotic activity is weakened at night. What do you think is the reason for this?

Diurnal animals are active during daylight hours. During the day, they spend a lot of energy moving and searching for food, while their cells “wear out” faster and die more often. In the evening, when the body has digested food, absorbed nutrients and accumulated a sufficient amount of energy, regeneration processes and, above all, mitosis are activated. Accordingly, in nocturnal animals the maximum mitotic activity of cells is observed in the morning, when their body is resting after an active night period.

*Tasks marked with an asterisk require students to put forward various hypotheses. Therefore, when marking, the teacher should focus not only on the answer given here, but take into account each hypothesis, assessing the biological thinking of students, the logic of their reasoning, the originality of ideas, etc. After this, it is advisable to familiarize students with the answer given.

1. Students must understand the concepts « mitosis, amitosis, meiosis, phases of mitosis, meiosis, spindle, karyokinesis, cytokinesis, homologous chromosomes, sister chromatids"; know the essence of the processes of mitosis and meiosis, the biological significance of different methods of cell division.

2. Students must be able to independently work through educational material, compare biological processes, give a reasoned answer, and draw conclusions.

3. Contributes b the formation of a scientific worldview and communicative qualities in students.

Equipment: tables on general biology “Mitosis”, “Comparison of mitosis and meiosis”, application “Methods of cell division”, presentation "Methods of cell division" computer, multimedia projector.

Progress of the lesson.

New cells can appear
only from previous cells.

I. UPDATED KNOWLEDGE, INTRODUCTION TO THE TOPIC.

A cell is not only a structural and functional unit of a multicellular organism, but also a unit of reproduction and transmission of hereditary information from generation to generation.

- What processes underlie the succession of generations?
- What is the essence of the cell life cycle?
- What structural organelles of the cell contain hereditary information?

In order for the resulting cells to be “full-fledged”, a complete set of genes must be transmitted. This is exactly what happens in most cases during cell division. Today we have to study in more detail the different methods of cell division and answer the question ( presentation “Methods of cell division”, slide 2 ):

II. STUDYING NEW MATERIAL.

The process of division is only part of the life of a cell. Between the two divisions there is a period called interphase (slide 3, 4)

Life cycle of a cell. Interphase (place in the life cycle, periods, main processes, meaning)– oral response, discussion.

During interphase, important processes occur: the synthesis of proteins and other necessary substances, the number of cellular organelles increases, and DNA replication. These processes are confined to different periods of interphase - G 1 (before doubling), S (DNA doubling), G 2 (after doubling) (slide 5) .

Interphase prepares the cell for division. There are several known ways of cell division (slide 6, 7 ) : amitosis, mitosis, meiosis.

The most common cell division is mitosis (slide 8) .

Mitosis - indirect cell division - oral answer, discussion.

Addition. An important role is played by centrioles, which form spindle consisting of microtubules (slide 9 ) . TO centromeric special proteins are attached to sections of chromosomes, forming a special structure - kinetochore. The spindle threads are attached to it. The number of threads can reach several thousand. From two to several dozen microtubules are attached to one chromosome. During anaphase, sister chromatids move toward the poles of the cell. This happens due to the shortening of the spindle microtubules and sliding against each other. Their gliding is ensured by proteins similar to muscle myosin. Microtubules are shortened due to rapid disassembly from the ends (+). It should be noted that in some organisms (protists and fungi), the nuclear envelope is preserved during mitosis (the spindle is formed inside the nucleus). In higher plants, there are no centrioles in the cells and the spindle is formed by other structures.

Cytokinesis occurs in several ways. In animals, cells lack a cell wall and are divided by a “constriction.” Even in anaphase, a ring of actin And myosin filaments. At the end of telophase, the ring begins to contract and draws the outer membrane inward. As a result, the cell becomes laced. The cytoplasms of higher plant cells are divided by a “septum”: vesicles from the Golgi apparatus are transported along microtubules to the cell equator. These vesicles begin to merge with each other in the central part of the cell, forming a small disk. Then new bubbles join it along the edges, and the disk grows. Formed median plate. A cell wall of cellulose begins to form on top of it.

Task 1 (worksheet). Scientists conducted studies of mitosis: it turned out that in animals that are nocturnal, in most organs the maximum of mitosis occurs in the morning and the minimum at night. In diurnal animals, the maximum is observed in the evening, and the minimum during the day. Analyze this fact.

Checking the completion of the task.

So, the meaning of mitosis is as follows (slide 10, 11) :

• Formation of genetically equivalent cells, continuity in a series of cell generations.
• Ensuring embryonic development and growth of organisms.
• Regeneration of organs and tissues.
• The basis of asexual reproduction of organisms.

In the process of evolution, a little later another method of division appeared - meiosis – reduction division. Its appearance is associated with early aromorphosis - sexual reproduction of organisms.

3. Meiosis - reduction cell division - oral answer, discussion.

Addition. The first division of meiosis is reduction, second - equational(there is no reduction in the number of chromosomes). The most significant changes occur in prophase 1. homologous chromosomes are located parallel to each other and touch, forming bivalents. Each bivalent contains 4 chromatids. At this time, a pair of chromatids from homologous chromosomes can exchange some sections. This exchange of DNA sections is called crossing over. As a result of the first division of meiosis, one homologous chromosome from a pair is transferred to the daughter cells. The resulting cells differ from each other in the set of genes: during crossing over, chromosomes exchanged sections and new gene combinations appeared. Thus, almost each of the resulting cells has a unique genotype, different from the genotype of the original cell (slide 12, 13 ) .

Task 2

(worksheet). Complete the drawings and look at them. Determine the effects of meiosis.

So, the features of meiosis are as follows (slide 16, 17 ) :

• During meiosis, daughter cells have half the set of chromosomes compared to the mother (in diploid organisms 2 n - haploid n).
• During meiosis, 2 consecutive divisions occur, in the interval between which DNA duplication does not occur. 4 cells (n) are formed.
• During the process of meiosis occurs crossing over.

4. Comparison of mitosis and meiosis

Completing task 3 (worksheet)

Task 3

(worksheet). Fill the table "Comparison of mitosis and meiosis".

Checking progress.

Let's try to answer the question posed at the beginning of the lesson

What is the mechanism for transferring the full set of genes from the mother cell to the daughter cells?

III. SUMMARY OF THE MATERIAL.

So, there are 2 main ways of cell division - mitosis and meiosis. Mitosis ensures the transfer to daughter cells of all the genetic material that was in the mother cell and equal distribution of this material between the daughter cells. During meiosis, the number of chromosomes in daughter cells is reduced by half compared to the mother cell. Mitosis ensures asexual reproduction, while meiosis ensures sexual reproduction of organisms.

IV. KNOWLEDGE CONTROL.

Test tasks (worksheet) – 5 – 7 minutes.

V. D\Z.

Prepare for the seminar “Forms of reproduction of organisms”, §§ 31 – 34 (students A.A. Kamensky, V.A. Kriksunov, V.V. Pasechnik. General biology, grades 10 – 11.

WORKSHEET

seminar "Methods of cell division".

Task 1. Scientists conducted studies of mitosis: it turned out that in animals leading a nocturnal lifestyle, in most organs the maximum mitoses occur in the morning and the minimum at night. In diurnal animals, the maximum is observed in the evening, and the minimum during the day. Analyze this fact.

Task 2. Finish the drawings and look at them. Determine the effects of meiosis.

Task 3.

Fill the table "Comparison of mitosis and meiosis".

Control tasks.

Level 1. Determine which pictures correspond to mitosis and which to meiosis.

Figure 4. Phases of mitosis and meiosis

Level 2

Option 1. There are organisms whose cells contain a triple set of chromosomes (triploid organisms - 3p). Meiosis in such organisms occurs with disturbances. Explain the cause of meiotic disorders in triploid organisms.

Option 2. There is a phenomenon in which, after chromosome replication, cell division does not occur - endomitosis(for example, in liver cells). What are the consequences of endomitosis? What biological significance might this process have?

Literature

1. A.A. Kamensky, V.A. Kriksunov, V.V. Beekeeper. General biology, grades 10 – 11. “Bustard”, M., 2005.
2. L.S. Vysotskaya, S.M Glagolev, A.O. Ruvinsky et al. General biology, 10 – 11 (textbook for classes with in-depth study of biology). "Enlightenment", M., 2001.
3. R.D. Petrosova, N.N. Pilipenko, A.V. Teremov. Didactic material on general biology. Belfarpost LLC, Minsk, RAUB-Citadel, M., 1997.
4. B.D. Commissioners. Independent and laboratory work in general biology. "Higher School", M., 1988.

Municipal budgetary educational institution

"Secondary school No. 6"

Safonovo, Smolensk region

elective course

for pre-professional training

Ostrovskaya E.I.,

biology teacher

MBOU "Secondary School No. 6"

Safonovo

2014

Biology in problems

What we know is limited

and what we don’t know is endless.

P. Laplace

Explanatory note

The program of the elective subject-oriented course “Biology in Problems” is intended for 10th grade students, is designed for 34 hours and can be implemented throughout the year. The program contains information and tasks that go beyond the scope of the general biology curriculum of the basic school, allows you to rethink the basic course, repeat and systematize the material covered, and decide on the choice of a natural and mathematical education profile.

Because In a biology course, not enough time is allocated for a full assimilation of difficult issues and the practical orientation of biological knowledge, then solving biological problems, conducting independent mini-studies and observations should contribute to the conscious assimilation of complex issues of the course and contribute to the development of the thinking skills of students who show an interest in biology.

Purpose of the program: expand students’ basic knowledge of biology and ensure conscious assimilation of the material through solving and composing biological problems of varying levels of complexity. Integrate acquired knowledge in biology, chemistry and mathematics. Carry out a professional test in the field of professions related to biology (medicine, genetics, ecology).

Program objectives:

Help students decide on a learning profile: identify abilities, inclinations, interests through solving biological problems;

Concretize, generalize and systematize theoretical knowledge in general biology;

Teach how to solve and compose biological problems based on acquired knowledge;

Consolidate and deepen knowledge of general biological patterns and terminology through solving and composing problems of varying levels of complexity and focus;

Develop students' thinking abilities;

Create a need to acquire new knowledge and ways to obtain it through self-education;

Develop the ability to conduct a scientific discussion, brainstorming, heuristic conversation;

As a result, students should:

decide on the choice of study profile in high school;

master the material at a qualitatively new level;

learn to solve and compose biological problems;

select practical material necessary for assignments;

apply your knowledge in non-standard situations;

use your knowledge, skills and abilities to solve practical problems.

Criteria for assessing mastery of course material:

observation of students’ activity when performing tasks of different levels of complexity and focus;

students’ self-assessment of completed tasks (reflective map);

survey of students at the end of the course.

Content

Topic 1. Introduction. Cytology. Unity in diversity. (4h)

Laboratory work: Studying cells of different organisms under a microscope. Comparative characteristics of cells. Analysis of the results obtained during observations, generalization and conclusions.

Practical work: Problem solving.

Topic 2. Secrets To summer metabolism (4h)

A distinctive property of living organisms is cellular metabolism. Catabolism and anabolism. Creation of supporting notes of the main processes.

Demonstration: computer presentation of protein biosynthesis processes.

Practical work: Problem solving. Preparation of tasks using acquired knowledge and reference material.

Topic 3. Basic instinct: methods of reproduction of organisms. Cell division. (4h).

Peculiarities of reproduction of organisms. Biological and ethical problems of cloning. Cellular engineering. Mitosis. Meiosis.

Demonstration: computer presentation of the basic processes of eukaryotic cell division.

Practical work: Problem solving. Drawing up clusters and diagrams of fission processes. Reading "blind tables".

Topic 4. Genetics. Patterns of Mendeleevian genetics. (6h)

Laws of Mendeleev's genetics: the rule of gamete purity, the law of dominance, the law of segregation, the law of independent segregation. Statistical patterns.

Practical work: Problem solving (mono-, dihybrid, analyzing crossing). Compiling problems using reference material and Mendel's laws.

Topic 5. Genetics. Is Mendel always right? (6h)

Basic provisions of Morgan's chromosome theory. Law of linked inheritance. Sex-linked inheritance. Medical genetic consultation, its goals and objectives. Basic forms of interaction of non-allelic genes.

Practical work: Problem solving. Drawing up chromosome maps. Role-playing game: medical genetic consultation. Drawing up a family tree based on the characteristics being studied.

Topic 6. Evolutionary doctrine. (2h)

Darwin's evolutionary theory: basic principles and criticisms. Basic provisions of the synthetic theory of evolution. Micro- and macroevolution. Main directions of evolution. When will evolution end?

Practical work: Problem solving. Tests. Preparation of tasks: “Find the error”, “Blind tables”, etc.

Topic 7. Ecology. Fundamentals of harmony in nature. (2h).

Laws and patterns of ecology. Biocenoses and ecosystems: composition, structure, properties. Biotic connections. Ecological pyramid rule. Population genetics. The Hardy-Weinberg law, its theoretical nature.

Practical work: Solving problems using the 10% rule and the Hardy-Weinberg law. Preparation of tests and tasks. Translation "from Russian into Russian".

Topic 8.Final lesson (1 hour) .

Game "Relay Race". Questioning students. Summarizing.

Thematic planning

№ p/p

Lesson topic

Total hours

Theory

Practical work

Forms of control

Introduction. Unity in diversity.

Laboratory work. Solving and composing problems

Secrets of Cellular Metabolism

Preparation of supporting notes. Solving and composing problems

Basic instinct: methods of reproduction of organisms. Cell division.

Problem solving. Drawing up diagrams of division processes. Reading "blind tables".

Genetics. Patterns of Mendeleevian genetics.

Solving and composing problems. Drawing up pedigree charts based on heritable traits

(Laws of linked inheritance. Interaction of non-allelic genes)

Solving and composing problems. Mini-research: “Genealogical tree of your family”

Role-playing game "MGK"

Problem solving. Tests. Compiling and filling clusters.

Solving problems, tests. Drawing up tasks. "Translation from Russian into Russian"

Final lesson.

Game: "Relay Race". Questionnaire

Literature

For the teacher:

1. Bodnaruk M.M. Biology. Additional materials for lessons and extracurricular activities. – Volgograd: Teacher, 2006

   Dmitrieva T.A., Gulenkov S.I. and others. 1600 tasks, tests and tests in biology for schoolchildren and applicants to universities. – M.: Bustard, 1999

   Kalinova G.S. etc. We pass the Unified State Exam. Biology. – M.: Bustard, 2007

   Kulev A.V. General biology. Lesson planning. – S.-P.: Parity, 2001

   Kulnevich S.V., Lakotsenina T.P. Quite an unusual lesson. – Voronezh: Teacher, 2001

   Murtazin G.M. Problems and exercises in general biology. – M.: Education, 1981

   Almost 200 problems on genetics. – M.: MIROS, 1992

For students:

Donetskaya E.G. General biology. Notebook with printed base (2 hours). – Saratov: Lyceum, 1997

Lebedev A.G. Biology. Exam Preparation Guide. – M.: AST, 2005

Ponamareva I.N. and others. Fundamentals of general biology. – M.: Ventana-Graf, 2006

Mamontov S.G. and others. Biology. General patterns. – M.: Bustard, 2002 - 2006

Annex 1

Educational materials

Topic 1. Introduction. Cytology. Unity in diversity. (2h)

Laboratory work

Comparative characteristics of cells

Target: consolidate the ability to work with a microscope, prepare microspecimens, study the structural features of cells of different organisms: find similarities and differences, draw conclusions.

Equipment: microscopes, micropreparations of plant, fungal and animal organisms, single- and multicellular, bacteria. Hay infusion, diluted yeast, coverslips and slides, glass rods, a glass of water.

Progress

Examine the proposed drugs under a microscope. Draw the cells, label the visible organelles.

Prepare a microslide of yeast fungi and bacillus. Draw the cells, label the visible organelles.

Find the main features of pro- and eukaryotic cells.

Compare cells of eukaryotes: plants, animals, fungi.

By what characteristics can eukaryotic cells be distinguished from prokaryotic cells using a light microscope?

Fill the table:

Distinctive features

Distinctive features of eukaryotes

Similarities between cells of all kingdoms of the organic world

Prokaryote

Eukaryotes

Plants

Mushrooms

Animals

Using the results of observations and knowledge gained in biology lessons, answer the questions:

1. Are all cells that do not contain a nucleus classified as prokaryotes? Why?

   Does the presence of chloroplasts always indicate that an organism belongs to the plant kingdom? Explain.

   Does the absence of chloroplasts always indicate that organisms do not belong to the plant kingdom? Explain.

8. Draw conclusions based on the results obtained.

Tasks

Complete the formula and identify the substance:

3. Complete the formula and identify the substance:

4. Complete the formula and determine the substance: C_ (H 2 O) n

5. Complete the formula and identify the substance: (- N H – CH – CO-) n

6. Complete the formula and determine the substance: N H ? – CH – COOH

On a fragment of one DNA strand, the nucleotides are arranged in the following order: A-A-G-T-C-T-A-C-G-T-A-T. Write a diagram of a double-stranded DNA molecule. What property of DNA was used to guide this? What is the length of this DNA fragment (the length of one nucleotide is 0.34 nm)? How many individual nucleotides (number and percentage) are there in this DNA?

Using the principle of complementarity, determine what nucleotide sequence the second strand of a DNA molecule will have if the nucleotide sequence of the first is as follows: -A-G-C-C-T-T-A-G-C-T-A-G-C-A-T -?

Using the principle of complementarity, determine what nucleotide sequence the mRNA will have if the sequence of the DNA chain template is as follows: -A-T-G-C-T-A-A-G-C-G-T-A-T-T-G -A-C-A-?

Using the principle of complementarity, determine what nucleotide sequence the section of the DNA molecule that served as the template for mRNA synthesis will have: - C-U-A-G-G-A-C-U-U-G-C-C-A-A-U- G-C-A-?

Determine the length of the mRNA chain, which contains 100 nucleotides, if the length of one nucleotide is 0.34 nm.

Determine the length of a section of a DNA molecule consisting of 500 nucleotides if the length of one nucleotide is 0.34 nm.

How many adenyl nucleotides are included in a DNA molecule consisting of 600 nucleotides, if thymidyl nucleotides make up 25%.

How many cytidyl nucleotides are included in a DNA molecule consisting of 600 nucleotides, if thymidyl nucleotides make up 25%.

How many guanyl nucleotides are included in a DNA molecule consisting of 800 nucleotides, if adenyl nucleotides make up 45%.

* The DNA macromolecule before reduplication has a mass of 10 mg and both of its chains have labeled phosphorus atoms. Determine what mass the reduplication product will have and why? How many and which strands of daughter DNA molecules will not contain labeled phosphorus atoms?

* A DNA molecule with a relative molecular weight of 69,000 is given, of which 8,625 are adenyl nucleotides. How many different nucleotides are there in this DNA and how long is it?

Note: The average relative molecular weight of one nucleotide is 345.

Reading "blind text"

Replace the highlighted phrases with the appropriate terms and fill in the missing words.

Complex substances consisting of simpler substances formed by a nitrogenous base, a five-carbon carbohydrate and one phosphoric acid residue, are called nucleic acids. They can be of two types: ... and .... DNA nucleotides contain sugar - .... and nitrogenous bases... types. RNA nucleotides contain sugar - .... and nitrogenous bases... types. Instead of a nitrogenous base... in RNA nucleotides there is.... Nucleotides in the chains of DNA and RNA molecules are connected by the most durable chemical communications. The chains in the molecule... are connected using hydrogen bonds according to the principle correspondences between certain nitrogenous bases of nucleotides. DNA and RNA molecules provide storage, transmission and implementation of hereditary information. Molecules...are made up of areas in which information about the primary structure of one protein is stored. There are... types of RNA: carrying information from DNA to the site of protein synthesis, transporting amino acids to the site of protein synthesis And supporting the structure of ribosomes.

Topic 2. Secrets of cellular metabolism . (2h)

Energy exchange

Three hares with different running speeds exhibit different rates of glucose oxidation and ATP formation in the muscles. Explain: a) how natural selection is expected to operate among these animals (all other conditions being equal); b) what is the role of heredity, variability and natural selection in improving the processes of energy metabolism?

When running a distance of 100 m, a person becomes hot and breathing quickens, but not immediately, but only after 50 m of running. Why?

Physiologists have found that the initial formation of a small amount of lactic acid in the muscles stimulates their contraction (for example, when warming up before running), and the accumulation of a large amount of lactic acid inhibits muscle contraction and causes rapid fatigue. In addition, during oxygen-free breakdown, a lot of glucose is consumed, and little ATP is formed. Explain what will happen to a person who has a weak heart if, during running or physical work, due to insufficient oxygen supply to the muscles, oxygen-free breakdown of glucose prevails in them. Give a scientific explanation for the accepted expressions “tired”, “didn’t have enough strength”.

* During the dissimilation process, 7 moles of glucose (C 6 H 12 O 6) were split, of which only 2 moles underwent complete oxygen split. Determine: how many moles of lactic acid (C 3 H 6 O 3) and carbon dioxide were formed? How many moles of ATP are synthesized? How much energy and in what form is accumulated in these ATP molecules? How many moles of oxygen are consumed to oxidize the resulting lactic acid?

SOLUTION: 1) out of 7 moles of glucose, 2 underwent complete breakdown, and 5 – incomplete;

2) create an equation for the incomplete breakdown of glucose

5 C 6 H 12 O 6 + 5∙ 2 H 3 PO 4 + 5∙ 2 ADP = 5∙ 2 C 3 H 6 O 3 + 5∙ 2ATP + 5∙ 2 H 2 O

3) create a summary equation for the complete breakdown of glucose

2 C 6 H 12 O 6 + 2∙6 O 2 + 2∙ 38 H 3 PO 4 + 2∙ 38 ADP = 2∙ 6 CO 2 + 2∙ 38ATP + 2∙ 44 H 2 O

4) find the total amount of ATP: 5∙ 2ATP + 2∙ 38ATP = 10+76= 86ATP

5) determine the amount of energy in ATP molecules: 86∙40= 3440 kJ

ANSWER: 10 mol of lactic acid, 12 mol of CO 2, 86 mol of ATP, 3440 kJ in the form of chemical energy in the macroergic bonds of ATP, 12 mol of oxygen.

* As a result of dissimilation, 5 mol of lactic acid and 27 mol of carbon dioxide were formed in the cells. Determine how many moles of glucose were consumed. How many of them have undergone only incomplete cleavage and how many have undergone complete cleavage? How much ATP is synthesized and how much energy is accumulated in them? How many moles of oxygen are consumed in the oxidation of lactic acid?

1) equation of incomplete breakdown of glucose

C 6 H 12 O 6 + 2 H 3 PO 4 + 2 ADP = 2 C 3 H 6 O 3 + 2ATP + 2 H 2 O

summary equation for the complete breakdown of glucose

C 6 H 12 O 6 + 6 O 2 + 38 H 3 PO 4 + 38 ADP = 6 CO 2 + 38 ATP + 44 H 2 O

3) from 1 mol C 6 H 12 O 6, 2 mol C 3 H 6 O 3 are formed. according to the conditions of the problem, 5 mol C 3 H 6 O 3 were formed, hence 5: 2 = 2.5 mol glucose

4) from 1 mole C 6 H 12 O 6 6 moles of CO 2 are formed according to the conditions of the problem, 27 moles of CO 2 were formed, hence 27: 6 = 4.5 moles of glucose

5) total glucose consumed: 2.5 + 4.5 = 7 mol

6) with incomplete breakdown of 2.5 mol of glucose, 2.5 ∙ 2 = 5 mol of ATP was formed

with the complete breakdown of 4.5 mol of glucose, 4.5 ∙ 38 = 171 mol of ATP were formed

7) a total of 176 moles of ATP were formed

8) 176∙40 = 7040 kJ of energy is stored in ATP

9) consumed for the oxidation of lactic acid: 6 ∙ 4.5 = 27 mol O 2

ANSWER: 7 mol; 2.5 mol of glucose underwent incomplete degradation, 4.5 mol of glucose underwent complete degradation; 176 mol ATP and 7040 kJ energy; 27 mol O 2 .

* Leg muscles, when running at an average speed, consume 24 kJ of energy in 1 minute. Determine how many total grams of glucose will be consumed by the leg muscles in 25 minutes of running if oxygen is delivered to the leg muscles in sufficient quantities to completely oxidize glucose. Will lactic acid accumulate in the muscles under such conditions?

SOLUTION: 1) in 25 minutes of running, 24∙25 = 600 kJ of energy will be consumed.

600: 40 = 15 moles of ATP release this amount of energy;

The molar mass of C 6 H 12 O 6 is 180, and from 1 mole of C 6 H 12 O 6, 38 moles of ATP are formed upon complete oxidation, hence: 180 ∙ 15: 38 = 71 g

ANSWER: 71g C 6 H 12 O 6; no, because completely oxidized.

Plastic exchange.

Compare three facts: 1 – protein molecules in the cell are constantly broken down as a result of dissimilation and replaced by new molecules of the same protein; 2 – protein molecules do not have the property of reduplication, therefore they cannot reproduce themselves; 3 – despite this, thousands of molecules of one type of protein newly formed in the cell are exact copies of the destroyed one. How do you think the synthesis of a large number of identical protein molecules occurs?

For protein synthesis in vitro, we took ribosomes and amino acids from spider cells, DNA and enzymes from Drosophila flies, and t-RNA and ATP from dogs. Whose proteins will be synthesized? Why do you think so?

The section of the DNA molecule encoding part of the polypeptide has the following structure: A-C-C-A-T-A-G-T-C-C-A-A-G-G-A. determine the sequence of amino acids in the polypeptide.

The sequence of mRNA triplets is given UGU-UAU-UUU-GAA-GAU-UGU-TsCU-TsGU-GGU, encoding the amino acid sequence in the protein vasopressin cis-tyr-fen-glu-asp-cis-pro-arg-gli. Determine: 1) how many nucleotides and triplets there are in mRNA; 2) what is the length of mRNA; 3) which triplets occupy the 3rd and 8th places in mRNA; 4) what nucleotides occupy the 5th and 21st places in mRNA; 5) how many amino acids are in the vasopressin protein; 6) how many types of amino acids are in the vasopressin protein; 7) which amino acid occurs twice and in what places?

On the left strand of DNA, the nucleotides are arranged in the following order: AGA-TAT-TGT-TCT-GAA. What primary structure will a protein synthesized with the participation of the opposite chain have? Which tRNAs will be involved in the biosynthesis of this protein?

* The primary structure of the hemoglobin protein normally has the following nucleotide sequence: val-gis-lei-tre-pro-glu-lyse. As a result of a gene mutation, Valine replaces glutamic acid, which leads to sickle cell anemia. Determine what changes in and where in the gene structure could lead to such consequences? What nucleotides can a codon encoding valine consist of?

* The protein molecule has the following sequence of amino acids: val-ala-gly-lys-tri-val-ser. Determine the structure of the section of the DNA molecule that encodes this polypeptide chain. Identify the anticodons of tRNAs involved in the synthesis of this protein. How will the structure of a protein change if the 5th and 13th nucleotides from the left are removed from the DNA region coding it? How will the structure of a protein change if in the DNA region coding it there is cytosine between the 10th and 11th nucleotides, and adenine between the 13th and 14th.

* An mRNA molecule is given: GAU-AUC-AUU-GGU-UCG. Determine: 1) the primary structure of the protein encoded in this molecule; 2) the number (%) of different types of nucleotides in the section of the DNA molecule that served as the template for the synthesis of mRNA; 3) the length of this gene; 4) the primary structure of the protein after the loss of the 9th nucleotide in the DNA chain.

How many nucleotides do genes contain, in which proteins are programmed, consisting of a) 10 amino acids; b) 25 amino acids; c) 150 amino acids?

* The molecular weight of the protein is 50,000. Determine the length of the corresponding gene.

11* . One of the DNA strands has a molecular weight of 34155. Determine the number of protein monomers encoded in this DNA.

Note: the average molecular weight of one amino acid is 100, the molecular weight of one nucleotide is 345.

Photosynthesis

Fill in the missing components of the reaction. Give a formulation for this process. Describe the main steps in this process.

6 CO 2 + 6 …→ C 6 H 12 O 6 + …

Can the dark phase of photosynthesis occur in the dark? Why?

During photosynthesis, only 1-2% of solar energy is converted into the energy of chemical compounds. What is the fate of the rest of the energy?

Currently, there is talk about the ecological aspects of photosynthesis. How do you understand this?

* During the day, one person weighing 60 kg consumes on average 30 liters of oxygen when breathing (at the rate of 200 cm 3 per 1 kg of weight per 1 hour). One 25-year-old tree (poplar) absorbs about 42 kg of carbon dioxide during the process of photosynthesis over 5 spring-summer months. Determine how many such trees are needed to provide oxygen to one person.

How are the problems of photosynthesis and food supply for the planet's population related?

Topic 3. Basic instinct: methods of reproduction of organisms. Cell division. (2h).

Think and explain: 1 – why, despite cell division, does the number of chromosomes remain constant in it? 2 – how is mitosis achieved by uniform distribution of chromosomes between daughter cells? 3 – what is the biological significance of mitosis?

Consider the diagram and determine how many chromosomes the cells will receive as a result of mitosis. Why do you think so?

Scheme of cell mitosis

M mother cell

Mitosis

Cell growth

Mitosis

Human cell Fly cell

In the nucleus of each somatic cell of a rabbit there are 22 pairs of chromosomes, and in Drosophila there are 4 pairs. Draw a diagram of how many chromosomes there will be in each daughter cell resulting from mitosis; meiosis.

* The total mass of all DNA molecules in 46 chromosomes of one somatic cell of the human body is 6∙10 -9 mg. Determine what the mass of all chromosomes in one daughter cell or two daughter cells formed by mitosis will be equal to. Why do you think so?

* Is it possible to determine which organism the tissue belongs to if a microslide is prepared from it so that the chromosomes in the cells are clearly visible? How can we explain this?

* Why do you think scientists call the metaphase plate a kind of passport of the body?

* Scientists conducted a study of mitosis: it turned out that in animals leading a nocturnal lifestyle, in most organs the maximum mitoses occur in the morning and the minimum at night. In diurnal animals, the maximum is observed at night, and the minimum during the day. Analyze and explain this fact.

It is known that in the house fly, each diploid cell contains 12 chromosomes. The second generation of flies (children) should have 24 chromosomes in each cell as a result of fertilization, and their grandchildren should have 48, etc. However, microscopic examination shows that in fact such an increase in the number of chromosomes does not occur in subsequent generations. In the cells of flies of any generation, the number of chromosomes remains constant - 12. How do you explain these contradictory facts? What adaptation has been developed that prevents an infinite increase in the number of chromosomes in the cells of individuals of the same species during sexual reproduction?

We present two contradictory facts:

1. During sexual reproduction, a young individual is formed as a result of fertilization, i.e. fusion of two germ cells;

2. In the cells of a young individual, the number of chromosomes does not increase and remains the same as in the parental individuals. As microscopic examination shows, the constancy of the number of chromosomes is preserved in grandchildren, great-grandchildren, and in all subsequent generations, although they are all the result of sexual reproduction.

Compare these facts and explain why during sexual reproduction, despite fertilization, the number of chromosomes in the cells of the offspring remains constant and does not increase with each subsequent generation.

In rabbits and female rabbits, each somatic cell contains 44 chromosomes. How many chromosomes do rabbits have in one egg? In one sperm? In the zygote? In the somatic cell of the offspring? How many sperm are produced from one primary germ cell (spermatogonia)? How many eggs are formed from one primary germ cell (oogonium)? Why?

Consider the diagram and determine how many and what chromosomes the daughter cells will receive.

Cell division diagram

Mitosis Meiosis

* Two groups of 100 diploid cells are given, each of them contains 8 chromosomes (A, A, B, B, C, C, D, E). In all cells of the first group, mitosis occurred, in the second - meiosis. Determine: 1) how many young cells were formed in the first group; b) how many and what chromosomes each cell of the first group contains; c) how many cells were formed in the second group; d) how many and what chromosomes does each cell of the second group contain?

* From two blastomeres formed from one zygote, two independent embryos developed and twins were born. How do you figure out what kind of twins they will be - identical or non-identical? Why?

* Two identical male twins married two identical female twins. Will the children from these marriages look like identical twins? Explain your answer.

As a result of abnormal mitosis in human cell tissue culture, the chromatids of chromosome 21 did not separate to different poles and ended up in one nucleus. How many chromosomes will the daughter cells have after this division?

As a result of abnormal meiosis in tissue culture of cat cells (38 chromosomes), the chromatids of the 15th pair of chromosomes did not diverge to different poles and ended up in one nucleus. How many chromosomes will the daughter cells have after this division?

As a result of abnormal mitosis in tissue culture of tomato cells (24 chromosomes), the chromatids of the 1st pair of chromosomes did not diverge to different poles and ended up in one nucleus. How many chromosomes will daughter cells have after such division?

1. Under the influence of various factors (chemical, temperature, etc.), sometimes during the process of meiosis in humans the divergence of homologous sex chromosomes is disrupted. Determine: a) which abnormal (according to the set of chromosomes) male gametes are formed when the divergence of sex chromosomes is disrupted during the process of meiosis; b) what abnormal zygotes (with a disrupted diploid set of chromosomes) are formed during the fertilization of normal eggs with abnormal sperm; abnormal eggs with normal sperm.

1. * American scientist J. Gurdon transplanted the nucleus from a frog cell into its egg, the nucleus of which was previously destroyed by ultraviolet irradiation. Thus, he managed to grow a tadpole, and then a frog identical to the individual from which the core was taken. What does this experience prove? Where is it currently used?

1. * In humans, there is a genomic mutation on the sex chromosomes, when cells contain only one X chromosome, that is, the chromosome set is not 46, but 45. Why are there no people with only a Y chromosome, without X? Explain your answer.

Reading "blind tables"

Phases of mitosis

Characteristic

Sequence of mitosis phases

Chromatids diverge to the poles, becoming independent chromosomes

Telophase

Metaphase

Spiralization of chromosomes occurs, the nucleolus and nuclear membrane disappear. Centrioles double

Methods of cell division

Characteristic

Biological significance

Cell duplication, genetic stability

Indirect division of a eukaryotic cell.

A method of dividing a eukaryotic cell that results in the number of chromosomes being halved.

Leads to a reduction in the number of chromosomes in gametes; ensures combinative variability and preservation of the karyotype of the species during sexual reproduction

Topic 4. Genetics. Patterns of Mendeleevian genetics. (3h)

Monohybrid cross

Determine how many and what kind of gametes the following genotypes form: AA, bb, Aa, AAVv, AaBv, AavvSs, AaBvSs, MmNnLlkk, GgDdWwSs.

What fruit shape is dominant in a tomato - spherical or pear-shaped - if in the first generation all tomatoes are spherical? What are the genotypes of the first and second generations? What Mendelian laws did you use to solve the problem?

In rabbits, black coat color is dominant over white. Can white rabbits be unclean? Why? What about the black ones? Is it possible to get black rabbits from crossing white rabbits? What about blacks? Why?

1) By crossing a white rabbit with a black rabbit, only black rabbits were obtained. Determine the genotypes of parents (P) and hybrids (F 1).

2) From crossing a white rabbit with a black rabbit, 6 black and 5 white rabbits were obtained. Determine the genotypes of parents (P) and hybrids (F 1).

When crossing guppies with gray bodies and guppies with golden bodies, offspring with gray and golden bodies were obtained. Is it possible to determine which gene is dominant and what are the genotypes of the parents?

* Gray rabbits were brought into the living area, considering them purebred. However, in the second generation, black rabbits appeared. Why?

When a pig with black bristles was crossed with a boar with white bristles, all piglets had black bristles. Write the genotypes of the parents and offspring.

In humans, the phenylketonuria gene is a recessive trait. From the marriage of healthy parents, a child was born with phenylketonuria. Determine the genotypes of the parents and the child.

What are the genotypes and phenotypes of the parent pair of guinea pigs if 2 of their offspring were smooth-haired and 6 were curly-haired?

Horned cows were crossed with a polled (hornless) bull. In the first generation, all calves were hornless. Determine the genotype of parents and offspring.

Crossing a white rabbit with a black female rabbit resulted in only black rabbits. Determine the genotypes of parents (P) and hybrids (F 1).

* Gray astrakhan fur (Shirazi) is more beautiful and more valuable than black astrakhan fur. Which sheep are economically profitable to select for crossing, gray and black Karakul lambs, if it is known that homozygous gray individuals are lethal (pure-bred lambs die in the first days after birth as a result of underdevelopment of the gastrointestinal tract).

In what ratio will the trait split in the offspring obtained from crossing a heterozygous gray Karakul ram with the same sheep?

Note: homozygotes for the dominant gene are lethal.

In humans, polydactyly (six-fingered) is a dominant trait. What is the probability of having a six-fingered child in a family where the mother is polydactylic and the father has a normal hand structure?

In humans, red hair color is a recessive trait. What are the genotypes of the parents, genotypes and phenotypes of the children, if the mother is red-haired and the father is non-red-haired and homozygous for this trait.

In tomatoes, the gene for red fruit color is dominant over the gene for yellow color. What is the phenotype and genotype of the first generation hybrids obtained from crossing alternative homozygotes? What phenotype and genotype will be the offspring from crossing first generation hybrids with each other?

Black coat color in dogs is dominant over brown. The black female was bred several times with the brown male. Each time the puppies were black. Explain why. What are the genotypes of parents and offspring? What kind of offspring can be expected from crossing these hybrids with each other?

What kind of offspring can be expected from the marriage of a blue-eyed man and a brown-eyed homozygous woman, if brown eye color is a dominant trait. Determine the genotype of the offspring.

In a person, the ability to use the right hand is a dominant feature. Can a family of right-handed homozygous parents have left-handed children? Why?

We crossed two night beauty plants with white and red flowers (incomplete dominance of red). Determine the genotypes of the parents, the genotype and phenotype of the first generation hybrids.

* When yellow-fruited raspberries were crossed with red-fruited ones, pink fruits appeared. How could this happen? Determine the genotypes of parents and offspring. What kind of offspring, in terms of phenotype and genotype, should be expected from crossing pink-fruited raspberry plants with each other?

When crossing long-fruited cucumbers with short-fruited ones, all hybrids of the first generation had an average fruit length. Why did this happen? What are the genotypes of parents and offspring? What phenotype and genotype will be the offspring from crossing first generation hybrids with each other?

When crossing tomatoes with pear-shaped and round fruits in the first generation, 1/2 of the offspring had pear-shaped fruits. Determine the genotypes of parents and offspring if the spherical shape of the fruit is a dominant trait.

The blood type depends on the action of three allelic genes A, B, O. They, when combined in pairs in human diploid cells, can form 6 genotypes: OO (I), AA or AO (II), BB or BO (III), AB (IV) . What blood types are possible in children if the mother has blood type I and the father has IV?

In the maternity hospital, two boys X and Y were confused. X has blood type I, and Y has blood type II. The parents of one of the boys have blood types I and IV, and the second one has I and II. Determine who is whose son.

A blue-eyed man, both of whose parents had brown eyes, married a brown-eyed woman whose father had brown eyes, like all his ancestors, and whose mother had blue eyes. Determine the possible genotypes and phenotypes of children from this marriage. Complete the pedigree chart.

R (grandparents)

F 1 (parents)

F 2 (children) F 2 F 2

26*. Make a chart of your family’s pedigree based on any hereditary trait (eye color, hair color, nose shape, lips, etc.)

27*. Write problems using Mendel's laws and reference materials for monohybrid crosses.

Dihybrid cross

What fruit shape is dominant in a tomato - spherical or pear-shaped, red or yellow in color, if in the first generation all the tomatoes are spherical, red-fruited? What are the genotypes of the first and second generations? What Mendelian laws did you use to solve the problem?

When crossing white rabbits with smooth hair with black rabbits with shaggy hair, the following offspring were obtained: 25% black shaggy, 25% black smooth, 25% white shaggy, 25% white smooth. Determine the genotype of the parents, offspring and type of crossing. White color and smooth coat are recessive traits.

When crossing two tomato varieties with red pear-shaped and yellow round fruits in the first generation, all the fruits turned out to be red round. Determine the genotypes of parents and first generation hybrids. What will be the ratio of phenotypes and genotypes of second generation hybrids?

* In cows, polledness (absence of horns) dominates over hornedness, and the gene for red coat color does not completely suppress the gene for white coloration, leading to intermediate inheritance of the trait - roan coloration. Determine the genotypes of the parents and the possible genotypes and phenotypes of the offspring if the bull is red horned and the cows are white hornless.

In humans, dark hair color (A) dominates light color (a), brown eye color (B) dominates blue color (c). Write down the genotypes of the parents, possible genotypes and phenotypes of the children born from the marriage of a fair-haired, blue-eyed man and a heterozygous brown-eyed, fair-haired woman.

* The absence of small molars and polydactyly in humans are inherited as dominant unlinked traits. Determine the genotypes and phenotypes of the parents and offspring if one of the spouses has small molars and is heterozygous for the polydactyly gene, and the other is heterozygous for the gene for the absence of small molars and has a normal hand structure. What is the probability of birth in this family of children with small molars and normal hand structure, and six-fingered children without small molars?

* The only child of nearsighted brown-eyed parents has blue eyes and normal vision. Determine the genotypes of all family members. What is the probability of having healthy children with blue eyes in this family?

* In humans, dark hair color dominates over light color, brown eye color dominates over blue. Determine the possible genotypes and phenotypes of children from the marriage of a fair-haired, brown-eyed woman, whose father had dark hair and brown eyes, and whose mother was fair-haired and blue-eyed, with a dark-haired, blue-eyed man, whose father was fair-haired and brown-eyed, and whose mother was dark-haired with brown eyes. Make a family tree.

As a result of crossing a pumpkin with spherical yellow fruits with a pumpkin with disc-shaped green fruits, all the first generation hybrids turned out to be disc-shaped yellow-fruited. When crossing hybrids of the first generation with each other in the second generation, 4 phenotypes were obtained: spherical yellow-fruited, disk-shaped yellow-fruited, disk-shaped green and spherical green. Determine the genotypes P, F 1 and F 2 using a crossing scheme.

R x

F 1

* Compose the problem according to the proposed scheme and solve it.

R X

F 1 X

F 2

Topic 5. Genetics. Laws of linked inheritance. (6h)

Sex-linked inheritance

In humans, the recessive hemophilia gene, like the dominant gene for normal blood clotting, is linked to the X chromosome.

1. Determine the genotypes and phenotypes of the offspring (sex and blood clotting pattern) from the marriage of a hemophilic man and a woman with normal blood clotting, homozygous for this trait.

   What is the probability of having a sick child in a family where both parents are healthy, but the woman’s father suffered from hemophilia?

By crossing a gray female Drosophila with a gray male, the offspring are gray females and all yellow males. How can we explain this? What are the genotypes of the parents?

In cats, the gene for red coat color and the gene for black coat color are sex-linked and are found only on the X chromosome. In addition, the interaction of these two allelic genes gives incomplete dominance - tortoiseshell coat color.

1. A tortoiseshell cat was crossed with a ginger cat. How will the splitting of hybrids proceed according to phenotype (sex and color)? Why?

   What will be the offspring of crossing a red cat and a black cat?

* In chickens, the dominant gene for silver and the recessive gene for golden feather color are localized on the X chromosome. In addition, in birds, the female sex is heterogametic, and the male sex is homogametic.

A silver White Wyandotte hen was crossed with a golden Leghorn rooster. Determine the ratio of phenotypes (by sex and color) in hybrids.

From parents with normal color vision, five children were born with normal vision and one boy was color blind (does not distinguish between red and green colors). How can we explain this? What are the genotypes of parents and children?

Chained inheritance

1. The distance between genes on a chromosome is measured in morganids (1 M = 1%). Determine the distance between genes in a chromosome if the frequency of recombined chromosomes is 15%.

1. The distance between the genes that determine the growth of a tomato plant and the shape of its fruit is 38M. Determine the frequency of recombined chromosomes.

1. Determine what genotypes and phenotypes the F 1 hybrids obtained by crossing a homozygous smooth-seeded pea plant with tendrils with a plant with wrinkled seeds and without tendrils will have. Both traits are localized on the same chromosome and the frequency of recombinant genes is 0%.

1. * The German researcher T. Morgan crossed a diheterozygous Drosophila with a gray body and normal wings with a male having a black body and rudimentary wings, i.e. carried out an analytical cross. However, instead of the expected 25% of each of the four hybrid phenotypes, he received: 8.5% gray with rudimentary wings, 8.5% black with normal wings, 41.5% gray with normal wings, and 41.5% black with rudimentary characteristics. Explain why this happened. Write the progress of the crossing and determine the distance between the genes in the chromosome.

1. Make a chromosome map of a tomato, using an arbitrary scale, if it is known that the distance between the growth genes (D - norm, d - dwarfism) and the fruit shape genes (O-round, o - oval) is 28M, between the leaf shape genes (L - norm , l – diseased leaf) and inflorescence shape (S – unleafed, s – leafy) – 17M, between growth genes (Dd) and leaf shape genes (Ll) – 73M, genes for fruit shape (Oo) and inflorescence shape (Ss) – 28M.

46* . Make a chromosome map of a tomato, using an arbitrary scale, if it is known that the frequency of recombined characters in hybrids: smooth fruit shape and weakly dissected leaves and ribbed fruit shape and strongly dissected leaves is 40%, smooth fruit shape and leafless inflorescence is 18%; leafy inflorescence and falling fruits – 2%.

Interaction of nonallelic genes

* Sweet pea has a dominant gene A causes the synthesis of a colorless pigment precursor - propigment. Dominant gene IN determines the synthesis of an enzyme, under the action of which pigment is formed from propigment. When these genes interact ( AB) sweet peas develop a purple color on the corolla. Two varieties of sweet peas with white flowers were crossed. In the first generation, all hybrids had purple flowers. Determine the genotypes P and F 1. How will the trait split into phenotype in the second generation?

* Dogs of the Cocker Spaniel breed with genotype A_B_ are black, with genotype A_bb - red, with genotype aaB_ - brown, with genotype aabb - yellow. A black cocker spanel was crossed with a light yellow one. A light yellow puppy was born. What color ratio should be expected from crossing this cocker spaniel with a dog of the same gentype?

* Pigs have a dominant gene G causes black coloration, and its recessive allele causes red coloration. However, if there is a dominant gene in the genotype R(repressor - suppressor) both of these genes are not expressed phenotypically.

When crossing black and white pigs of different breeds, only white pigs appear in the first generation. Crossing them with each other leads to the appearance of white (12/16), black (3/16) and red (1/16). Write the genetic diagram of the crossing.

* The skin color of Negroid representatives is determined by two pairs of genes AABB, skin color of whites - by their recessive alleles. Mulattoes have varying degrees of intermediate skin color. Determine the skin color of the children from the marriage of a black man and a mulatto woman ( AaBb)

Topic 6. Evolutionary doctrine. Reasons and patterns of diversity and development of living nature.(4h)

1. In the meadow grow meadow timothy, wheatgrass, meadow bluegrass, red clover, pink clover, meadow rank, mouse pea, peach bell, and bluebell. Determine how many species and genera grow in the meadow.

1. Two cultivated plants, barley and rye, have the same number of chromosomes (14), but do not interbreed and differ in appearance and chemical composition. Determine: 1 – these plants belong to the same or different species; 2 – what species criteria were you guided by?

1. * In two lakes that do not communicate with each other, freshwater fish live: crucian carp, roach, ide, bream, pike perch. Determine: 1 – how many fish populations live in the first lake; 2 – how many fish populations live in the second lake; 3 – how many species of fish live in two lakes; How many fish populations live in the two lakes?

1. * Brown hares live both in the steppes of the Trans-Urals (east of the Ural Mountains) and in the steppes of the Cis-Urals. They are separated by the mountain forests of the Urals (geographical isolation), but are outwardly indistinguishable and, when crossed, produce fertile offspring. Determine what form of existence of the species these hares constitute:

a) one population of one species;

b) two populations of the same species;

c) one population of two species;

d) two populations of two species.

1. Determine the form of variability of organisms: modification, mutation or correlative

The feeding of the cows has been improved on the farm, resulting in an increase in milk yield;

In a brood of jackdaws, one chick turned out to be an albino;

A short-legged lamb was born from a ewe with normal legs, from which a new breed of sheep later appeared - the Ascona.

On well-fertilized soil, cabbage produces large heads of cabbage, and on poor soil, small heads;

A hairless puppy with underdeveloped teeth was born;

Dogs living outside in winter have thicker fur than indoor dogs;

The baby crane has a longer beak and legs than other chicks;

The rock pigeon has a chick with feathered legs and webbed toes;

In primrose, one flower turned out to be larger than the others and had six petals;

The dog has developed a conditioned reflex to offer its paw.

As a result of the drought, the wheat harvest was less than expected;

One of the shoots of the fragrant tobacco plant has striped leaves.

6. Determine the forms of the struggle for existence, write down the corresponding numbers in the table:

Plant seeds are digested in the digestive tract of animals;

The hare eats the bark of fruit trees;

A man cuts down a forest;

Layering in a mixed forest;

Thinning of the birch forest as a result of lack of light;

Dandelion seeds fell into the lake;

The cuckoo threw the flycatcher eggs out of the nest;

Potato yields have declined as a result of late blight;

The person fell ill with the flu;

Hyenas eat lions' scraps;

Acacia seeds fell on a sand dune;

In the spring, the deer rut begins;

In winter, wolves form packs.

14) The storm washed the starfish ashore.

Intraspecific struggle

Interspecies fight

Fighting adverse conditions

The number of hares in the central regions of the European part of the USSR in 1932. It was 2% (compared to the number in 1959), 1938-30%, 1941-75%, 1944-8%, 1948-2%, 1950-10%, 1952-100%, 1954-70%.

a) Construct a graph and determine which elementary evolutionary factor this example illustrates;

b) How, in your opinion, does a change in the number of hares affect the number of their enemies - lynxes, foxes, wolves?

8. One dandelion plant occupies an area of ​​10 m2 on the ground and produces about 100 flying seeds per year. Determine: 1 – How many square kilometers of area will be covered by all the offspring of one dandelion individual in 10 years, provided that it survives exponentially and all individuals survive. 2 – Do plants have enough space on the surface of the earth’s land for the 11th year of reproduction? 3 – Why doesn’t this happen?

Note : land surface area 148 million km 2

ANSWER: 1 –10 12 km 2 ; 2 – no; 3 – various forms of struggle for existence operate.

10* Three individuals of the bluebell had different fates. One plant was eaten by caterpillars even before flowering (they were attracted by leaves and stems that were more tender and juicy than the others); the other had an inconspicuous corolla without aroma, and therefore did not attract pollinators and did not leave fruits or seeds; only the third produced full-fledged seeds.

What property of the body led to such consequences? Which individuals should be considered “losers” in the struggle for existence? Which plant died as a result of antibiotic relationships between organisms, and which as a result of a violation of symbiotic relationships?

In the winter of 1898, after heavy rain and snow, Brown University researcher H.C. Bumpus collected and brought to the laboratory 136 house sparrows stunned by the elements. Of these, 72 survived and 64 died. Bumpus measured the total body length, wingspan, body weight, beak and head length, humerus and femur length, skull width, and keel length for all individuals. His measurements showed that in surviving birds all these characteristics are close to average values. The result of what natural selection did the scientist reveal?

In the reserve, the grass was constantly mowed and hay was dried to feed the animals in the winter. As a result, two races of rattles formed on the territory of the reserve: spring rattle and autumn rattle. Explain what form of selection this example demonstrates and what it might lead to?

Human cells of any race contain 46 chromosomes, fertile offspring are born from interracial marriages, the blood of a person of one race can be transfused to people of other races if the blood groups and Rh factor match. What conclusions can be drawn from these factors? What criteria do the factors listed above apply to?

Compare the behavior of social insects (bees, ants), a herd of monkeys and human society. Explain: 1) in the life of which of them biological and social factors of evolution operate, prove; 2) in the life of which of them biological and social laws operate, prove.

15.* Several interrelated biological phenomena and their results are listed:

1. Mutational variability;

   Modification variability;

   Heredity;

   Artificial selection;

   Formation of new breeds (varieties);

   Correspondence of breeds (varieties) to human interests;

   Variety of breeds (varieties);

   Human needs to obtain the necessary qualities from domestic animals.

ABOUT identify and depict schematically, with the participation of which biological phenomena various breeds (varieties) occurred and what results this led to.

16.* Imagine that you are a breeder and have at your disposal only one pair of pigeons - wild rock pigeons. You are faced with a problem: to breed from them a new breed of pigeons with black plumage. How will you do this? Propose a plan of consistent actions for 5 years. Note : Crossing with other breeds of pigeons is not allowed.

17.*H. Darwin, during his trip around the world (1831 - 1836), studied in South America the life of semi-wild indigenous people - Indians, forced out by white colonialists on the island. Tierra del Fuego. Darwin writes about them that during any famine, savages always keep several of the best dogs for the tribe. Determine: 1 – what form of selection we are talking about; 2 – what result did long-term selection among dogs lead to? Why? 3 – what role does the heredity and variability of dogs play in preserving their qualities useful to humans?

Despite the intense struggle of people against rats and house mice, they have not yet been exterminated. Explain whether selection currently occurs among rats and mice. What kind of selection is this?

When the antibiotic penicillin began to be used, it was the most reliable medicine against lobar pneumonia. However, this did not last long. Now even large doses of penicillin have no effect on the bacteria that cause this disease. Explain the reason for this phenomenon.

We list several biological phenomena studied by Charles Darwin:

1) mutational variability;

2) modification variability;

3) heredity;

4) artificial selection;

5) divergence;

6) the formation of several new breeds (varieties) from one parent species;

7) adaptability of breeds and varieties to human interests;

8) variety of breeds and varieties;

9) human needs to increase the productivity of domestic animals and cultivated plants;

10) struggle for existence;

11) natural selection;

12) the formation of several new subspecies and species from one parent species;

13) the relative fitness of an organism for the benefit of a population or species in the wild;

14) diversity of species in nature;

15) gradual complication of organisms in nature.

Determine and depict schematically, with the participation of which phenomena listed above, various species occurred, for example, tits, and what results this led to. Show the relationship between these phenomena with arrows, directing them from causes to effects. b) What is the role of each of these phenomena? c) In the diagram, circle in a double circle the phenomenon that relates to the main driving force (factor) of evolution in living nature.

Determine and depict schematically which of the phenomena listed below contributed to the long legs and neck of the giraffe: show the relationship of these phenomena with arrows, directing them from causes to effects.

Phenomena: 1) mutational variability; 2) modification variability; 3) heredity; 4) artificial selection; 5) divergence; 6) the formation of several new breeds (varieties) from one parent species; 7) adaptability of breeds and varieties to human interests; 8) variety of breeds and varieties; 9) human needs to increase the productivity of domestic animals and cultivated plants; 10) struggle for existence; 11) natural selection; 12) the formation of several new subspecies and species from one parent species; 13) the relative fitness of the organism for the benefit of the population or species in the wild; 14) diversity of species in nature; 15) gradual complication of organisms in nature.

22. It is known that many types of microorganisms are able to quickly adapt to changing environmental conditions. What is the mechanism that ensures the high adaptability of microorganisms?

There are several populations within species X. From population A, which had advantages, subspecies A 1 gradually emerged. What is this evolutionary process called? What elementary evolutionary factors are involved?

Subspecies A1 became increasingly isolated and gradually stopped interbreeding with other populations of species X, resulting in the emergence of a new species Y. What is this evolutionary process called? What elementary evolutionary factors are involved?

Compare two adjacent types of organisms and explain what phenomenon their similarities or differences relate to: convergence or divergence.

Mole cricket (insect) and mole (similarity in the shape of the front legs).

Stiffleaf buttercup and golden buttercup (difference in structure).

White hare and brown hare

Camel and fat-tailed sheep (fat reserve)

Polar bear and brown bear

Shark and dolphin

Grape snail and pond snail

Crayfish and crab (have claws)

Kangaroo and ostrich (long hind limbs)

Frog and Toad

Flying lizard and bat (wings)

Whale and fish (body shape).

* Two points of view are offered:

1. Adaptability in the structure and behavior of organisms of any species in the process of evolution has already reached perfection and speciation no longer occurs, because natural selection has already “managed” to improve everything over billions of years.

2. Any modern species has its drawbacks, in addition, the environment is constantly changing, so natural selection always occurs where there is life.

Express and give reasons for your opinion about the place and role of natural selection at the present stage of the development of life on Earth.

* A discussion arose between students in a biology lesson.

One student argued: the fitness of species is an indisputable universal fact. It is explained by the fact that any living organism responds to any change in environmental conditions, albeit unconsciously, with an adequate change in its organs and functions, because adequate variability is an innate ability of organisms that arises from the first days of life.

Second student: All organisms from the moment of the emergence of life on Earth have variability as a universal property of living matter. But not a single organism has ever possessed or possesses the original property of only adequately changing under the influence of environmental conditions. And modern organisms just cannot change adequately.

Third student: objects to the second student's last statement. And he believes that modern organisms have already acquired, as a result of natural selection, the property of adequate (adaptive) variability.

Fourth student: modern organisms, if environmental conditions change, can respond with temporary adaptive changes in some of their characteristics, but such a reaction, like any other adaptation, arose as a result of natural selection. However, modern organisms have not acquired adequacy of variability as a property.

Analyze student statements. Express and justify your point of view.

28. Determine the main directions of evolution, fill out the table:

Aramorphosis

Idiomatic adaptation

General degeneration

Adaptive characters that arose during evolution:

The emergence of multicellularity;

Spine formation;

The appearance of flippers in seals;

Formation of a 3-chambered heart in amphibians;

The appearance of a climbing stem in grapes;

Loss of chlorophyll in broomrape;

Absence of pistils and stamens in reed sunflower flowers;

Formation of the trunk in an elephant;

Reduction of eyes in a mole;

The appearance of spines on a cactus;

The emergence of photosynthesis;

Loss of the digestive system in the bovine tapeworm.

29.* The found paleontological remains of a mammoth contain 5.25% of radioactive carbon (14 C) of its original amount in animal tissues. Determine the geological age of the mammoth using a carbon clock.

Note: The half-life of 14 C is 5360 years. The accuracy of age determination is not absolute ± 3% of the calculated age.

SOLUTION: 1. We take the initial amount of 14 C content as 100%, hence: 50% - 1 half-life; 25% - 2 half-lives; 12.5% ​​- 3 half-lives; 6.25% - 4 half-lives. Thus, 4 full half-lives of 14 C: 4 ∙ 5360 = 21440 years.

Find the remaining %: 6.25 – 5.25 = 1%

Find the time as a result of which the content of 14 C decreased by 1%

6.25: 2 = 3.125% - 5th half-life, i.e. 5360 years

3.125% – 5360 years

1% – X years

X = 5360: 3125 = 1715.2 years

4. General age: 21440 + 1715,2 = 23155.2 years (±695 years)

30.* The content of radioactive carbon in the found paleontological remains is given: a) ancient deer - 12%; b) ancient horse – 6%; c) ancient bull – 3%.

Determine the geological age of these animals using carbon clocks.

Note: half life 14 C is equal to 5360 years. The accuracy of age determination is not absolute ± 3% of the calculated age.

Answer. a) 16,500 (± 495) years; b) 21.870 (± 656) years; c) 27.230 (± 817) years.

Topic 7. Ecology. Fundamentals of harmony in nature. (4h)

What abiotic factor turned out to be the main regulator and signal of seasonal phenomena in the life of plants and animals in the process of evolution? Why this particular factor and not another?

Low temperatures limit the distribution of moose in Scandinavia and Siberia. Although the average annual temperature in Siberia is higher, moose are found much further north in Scandinavia than in Siberia. Explain why.

In Yakutia, Daurian larch grows on the northern slopes, and the southern slopes are covered with pine forests. Explain this distribution of trees.

The fennec fox lives in the deserts of Africa and has very large ears, the common fox, characteristic of temperate forests, has medium-sized ears, and the arctic fox, which lives in the tundra, has very small ears. Explain why these systematically close species differ significantly in the size of their auricles.

Sometimes in the summer in the morning, after a cool rainy night, many plants show signs of wilting, although the soil is very moist and the air temperature is quite high. Explain the reasons for plant wilting.

B X I X century German physiologist K. Bergmann established a zoogeographical pattern: the body size of warm-blooded animals in the Northern Hemisphere increases when moving north, and in the Southern Hemisphere - when moving south. What explains this phenomenon?

Many desert members of the cactus family in the New World and the euphorbia family in the Old World are characterized by thick stems that store water and prickly leaves that protect them from being eaten by animals. Why did plants belonging to different families and growing in different parts of the world develop similar characteristics?

The body of sculpin, trout, and minnow is almost round in cross section, while the body of roach, perch, and crucian carp is laterally compressed. What is the reason for the differences in body shape in these fish?

It is known that the process of fertilization in flowering plants occurs at fairly high temperatures. How does the inside of the flowers of alpine and arctic plants reach temperatures higher than the ambient temperature?

Draw up a diagram of the food chains of an aquarium containing crucian carp and guppies, pond snails and spool snails, elodea and vallisneria plants, slipper ciliates, and saprophytic bacteria.

Consider the food chain: cereals → grasshopper → frog → snake → snake eagle. Using the rule of the ecological pyramid, build a pyramid of biomass, based on the fact that during the period of development of the snake-eagle, its mass is 5 kg.

Based on the rules of the ecological pyramid, determine how much plankton is needed so that 1 dolphin weighing 400 kg can grow and exist in the Black Sea.

It has been established that among insects the highest fertility is in herbivorous forms, and the lowest in predators. Explain why.

It has been established that at the edges of forests and in transition zones (for example, forest-steppes) there is a high species diversity and a higher population density of living organisms than in adjacent biocenoses. Explain this phenomenon.

It has been established that in tropical forests there is never an outbreak of the number of individual species, while the tundra is characterized by mass reproduction of lemmings, sharp fluctuations in the number of arctic foxes and other animals. Explain why in tropical forests there are no sharp fluctuations in the numbers of individual species, but in the tundra such phenomena are natural.

Life in sandy deserts is richer than in clay deserts. Plants here reach large sizes, and soil animals have greater species diversity and greater numbers. Explain the reasons for the greater diversity of life in sandy deserts compared to clay deserts.

In the north, the stylized forms of dwarf birch, spruce, juniper and cedar have upper branches that rise high above the ground, usually half-dead or dead, while creeping ones are alive. Explain the reasons for this phenomenon.

Ornithologists have found that three species of warblers that live in the same forests feed on insects on different parts of the trees. Blackburn's Warbler feeds in the upper parts of tree canopies, Chestnut Tree Warbler feeds in the middle of canopies, and Yellow-headed Tree Warbler forages in lower parts of tree canopies. Explain the reasons for different feeding sites among closely related species of warblers.

Blackburn's Warbler Chestnut Yellow-headed Tree Warbler Tree Warbler

In many groups of organisms, asexual reproduction is common in the summer, and in the fall, with lower temperatures and shorter daylight hours, a transition to sexual reproduction occurs. Laboratory experiments have shown that by changing environmental conditions: depriving food, heat, light, oxygen, etc., it is possible to force organisms to switch to sexual reproduction in the summer. What is the biological meaning of the alternation of sexual and asexual reproduction in organisms?

Research has established that on every square meter of small cabbage fields there are on average up to 69 cabbage white caterpillars, and on one square meter of large fields no more than one caterpillar was found. At the same time, pests in both large and small fields are concentrated mainly in the edge zone, 30-40 m wide. Similar results were obtained when taking into account the population density of other pests: cruciferous flea beetle, clover seed eater, apple codling moth, etc. Why is the number of insect pests of agricultural crops significantly higher at the edges of agrocenoses and in small fields? What measures can be recommended to reduce the degree of damage to crops by phytophagous insects, taking into account the characteristics of their distribution?

Can a shelterbelt forest planted by man exist independently if it consists of only one type of tree? Why?

How and why will the life of an oak grove change if: a) all the bushes are cut down; b) chemically destroyed herbivorous insects?

With the mass shooting of birds of prey that exterminate partridges and black grouse, the latter die out in the forest; when wolves are destroyed, deer die out; As a result of the destruction of sparrows, the grain harvest falls. How can we explain this?

Many animals store seeds and fruits of plants for the winter. For example: a pair of yellow-throated mice stores 38,000 beech nuts in two weeks, a wood mouse can store up to 15,000 acorns in 6 days. Explain how plant populations are influenced by animals that store their seeds and fruits for food.

On average, a bumblebee family consists of 100 working individuals, making at least 20 flights on a fine day. For each flight, one bumblebee visits about 240 flowers. How many flowers can a family of bumblebees pollinate in one month? What is the importance of bumblebees in the evolution and ecology of insect-pollinated flowering plants?

It has been established that a worker bee visits up to 12 flowers in 1 minute, and about 7,200 per day. In a strong colony there are up to 50,000 worker bees (in a weak colony there are about 10,000). How many flowers can bees of one colony pollinate in one day? What is the practical and biological significance of bees?

The males of many animals, especially during the breeding season, guard a certain territory, into which they do not allow any individual of their species, except their females. Explain the significance of this territorial behavior.

Many species of animals normally exist only in fairly large groups: bacteria can only live in colonies numbering at least 10,000 individuals, African weavers - in conditions where there are at least three nests per 1 m 2, the most productive herds of reindeer include 300-400 individuals. Explain why some animal species develop normally only when combined into fairly large groups.

Find the mistake and justify your choice:

Polar bears do not eat penguins because their tough meat is not tasty.

Pigeons feed their chicks with bird's milk.

Crocodiles are aquatic animals.

Birds of prey have powerful claws and teeth.

When the dominant male in a pride changes, the new male kills all the cubs.

If only female or male guppies live in the aquarium, then no offspring can be expected.

Scientists have found that coniferous trees are more damaged by industrial gases than deciduous trees. Explain why.

The total amount of oil and oil products that enter the waters of the World Ocean annually exceeds 10 million tons. How do oil films affect the exchange of substances between the ocean and the atmosphere? What effect do petroleum products entering the ocean have on the life of living organisms?

In California, in order to kill mosquitoes, the water in Clear Lake was treated with DDT at a concentration of 0.02 mg/l. After some time, the fish-eating birds living on this lake stopped hatching chicks due to the non-viability of the embryos. Explain what is the connection between the treatment of water in the lake with DDT and the non-viability of bird embryos from Lake Clear?

For the manufacture of aerosol cans with medicines, cosmetics, and household products, freon gas is used, which does not have a harmful effect on organisms. However, scientists insist on limiting the use of this gas. Why?

To protect plants from pests, pesticides are now widely used in almost all countries of the world. How does this affect human health and the environmental sustainability of local biocenoses? Give reasons for your answer.

A dam is planned to be built on a river flowing through European Russia. Suggest possible changes in the ichthyofauna in this river.

In the 80s X IIn the 10th century, all citrus plantations in California almost died due to the grooved Australian mealybug, sucking plant juices. This pest is random was brought to America from Australia. After the methods used to combat this pest did not produce results, 129 specimens of the mealybug’s natural enemies, the predatory rhodolia beetles, were brought from Australia. In the spring of 1889, 10,000 rhodolia were released into California orange plantations, and by October of that year, the Australian grooved bug had been literally eradicated from most of southern California. The biological control method has proven to be quite effective and has been used for more than 50 years. However, the use of the pesticide DDT allowed the scale insect population to flourish again. Why did using a pesticide in pest control backfire?

Determine the frequency of occurrence of the recessive allele in a population where dominant homozygotes (HH) are 81%.

What is the proportion of heterozygous individuals in an equilibrium population where recessive homozygotes make up 64%?

SOLUTION: According to the Hardy-Weinberg law, in an equilibrium population the ratio of dominant and recessive genes is 1: p + q = 1

According to the conditions of the problem, individuals with a genotype ah (q 2) are 64% (0.64), which means the frequency of occurrence of the allele a: a = √ 0,64 = 0,8,

Then, the frequency of occurrence of the allele A: p = 1 – 0.8 = 0.2

Consequently, the proportion of heterozygous individuals in a given population, in accordance with the Hardy-Weinberg law: Ahh = 2рq = 2 ∙ 0.8 ∙ 0.2 = 0.32 (32%).

ANSWER: the proportion of heterozygotes is 0.32.

* Z-Hardy-Weinberg. There is a population of the following composition: 0.49 AA , 0,2 Ahh , 0,09 ah . Is the viability of individuals with different genotypes the same?

SOLUTION: If the viability of all individuals is the same, then the population must be in equilibrium and allele frequencies must satisfy the Hardy-Weinberg law: p 2 + 2 pq + q 2 = 1, p + q = 1

(AA + 2Aa + aa = 1, A + a = 1)

Because genotype frequency AA 0.49, a ah 0,09,

then the allele frequency A = √ 0,49 = 0,7

A = √0,09 = 0,3

This means that the frequency of occurrence of heterozygotes should be:

Ahh = 2 ∙ 0,7∙ 0,3 = 0,42

And according to the conditions of the problem, heterozygotes are 0.2, therefore the viability of heterozygotes is reduced.

Translation from Russian into Russian

What famous Russian proverbs and sayings are “hidden” in expressions?

1. Lost its azimuth in a small biocenosis consisting of gymnosperms.

(Lost in three pines).

1. A type of socially useful human activity, without which it is not possible to remove a small component of the fauna of a standing aquatic biocenosis.

(You can’t even pull a fish out of a pond without difficulty).

1. A representative of the class of mammals, which, despite good nutritional conditions, constantly strives to join the arboreal community.

(No matter how much you feed the wolf, he constantly looks into the forest).

1. An imprudent action, as a result of which a male small ruminant fell into a vegetable agrocenosis.

(They let the goat into the garden).

1. An unpleasant sensation caused by typical representatives of the carnivorous order of the middle zone, which interferes with visiting the arboreal community.

(If you are afraid of wolves, do not go into the forest).

1. The hydrophobic effect is well represented in representatives of waterfowl of the order Anseriformes.

(Like water off a duck's back).

Translate proverbs and sayings from Russian into Russian using biological terminology:

The word is not a sparrow; if it flies out, you won’t catch it.

Work is not a wolf; it will not run away into the forest.

That's why the pike is in the sea, so that the crucian carp doesn't doze off.

The dog barks - the wind blows.

The forest is being cut down and the chips are flying.

Rotten apple injures its neighbors.

Mini-studies

Topic 5. Genetics. Is Mendel always right?

Drawing up a pedigree based on the trait being studied

Target :

Study your family's pedigree;

Learn to use genealogical symbols when compiling a family pedigree;

- divorceTrace how your chosen trait is inherited in your family tree.

1. 1. 1. Determine the nature of inheritance of the trait (dominant, recessive, sex-linked, autosomal).

1. 1. 1. Determine the patterns of inheritance of the selected trait.

         Draw conclusions about the nature and statistics of inheritance of the trait.

Topic 6. Evolutionary doctrine. Reasons and patterns of diversity and development of living nature.

The influence of environmental factors on the variability of organisms.

Target: Identify the influence of environmental factors on the variability of organisms.

Equipment: mucor, nutrient substrate, dissecting needle, glass vessels.

Progress

Determining the effect of light on variability

mold fungus mucor.

1. 1. Grow mucor (white mold) on damp bread or boiled beets (carrots), which you place in a glass or plastic bag and place in a warm place.

1. 1. After the mucor spores have matured, take two glass vessels, place a nutrient substrate (bread, vegetables) at the bottom of which, and use a needle to transfer some mucor spores there.

1. 1. Close the jars. Place one vessel (experiment) in a dark place, leave the other (control) in the light. All other factors (temperature, humidity, etc.) should be the same. After 7 days, compare the results: the density of mushroom filaments and sporangia. Record the results in the table:

Conditions of experience

Density of hyphae

(where it’s more magnificent)

Density of sporangia

(where more)

Conclusion

In the dark

What variability was evident in this experiment?

Topic 7. Ecology. Fundamentals of harmony in nature.

Study of forms of relationships between organisms.

Target: Study the forms of biotic relationships in the ecosystem and the influence of abiotic factors on the viability of organisms.

Equipment: nutrient substrate, mucor, penicillium, dissecting needle, Petri dishes.

Progress

1. 1. 1. 1. 1. 1. Grow mukor in three jars with bread (vegetables). At the same time, in a separate jar, grow penicillium (green mold) on the bread.

                  Infect the flour in one jar with penicillium spores, place another jar nearby without infecting it with penicillium, and place the third in a hot, dry place. Close all jars.

                  Observe the processes occurring in the tanks for 7–8 days.

                  Explain the reasons for what is happening.

Conditions of experience

Observation results

Conclusion

Mucor with penicillium

Mucor without peni-zilla

Flour in a hot dry place

Topic 5. Genetics. Is Mendel always right?

ROLE-PLAYING GAME

Medical genetic consultation

Target:

1. Introduce students to the purpose of medical genetic consultations;

2. Show the importance of the laws of genetics in the practical life of a person;

3. Improve skills in composing and solving genetic problems.

Organization: 3 groups, mutually changing roles: patients, geneticists (medical consultants), experts.

Role of patients : draw up tasks that require contacting a medical genetic consultation.

Role of consultants : solve patients’ problems, determine possible options for the development of events.

Role of experts: will evaluate the correctness and complexity of tasks compiled by “patients” and the completeness of the solution to the problem by “consultants”.

Equipment: badges, reference materials on human genetics, peer review sheet.

Lesson plan

Teacher's opening speech.

Role-playing game "MGK".

Summarizing.

Progress of the lesson

Teacher's opening speech.

Genetics claim that 95–98% of diseases known to mankind are hereditarily predisposed: these are defects of internal organs, physical deformities, mental illnesses, diseases associated with metabolic disorders, etc. Therefore, the main task of medical genetic consultation is to determine the likelihood of having a sick child in a family where at least one of the parents or relatives suffers from a hereditary disease. Currently, due to the deterioration of the environmental situation, the role of MGC is increasing. Some chemical materials used in everyday life, cosmetics and perfumes, medicines, food additives, genetically modified products, various types of radiation, as well as alcohol, tobacco, drugs, etc. – all of these are powerful mutagens that increase the level of mutations in the human body. Somatic mutations that occur in humans are not inherited and may not even appear if they are recessive. But generative mutations are dangerous because healthy parents can give birth to children with serious hereditary diseases. Realizing this, many families who want to have healthy children turn to medical genetic consultations, where, as a result of laboratory tests, using modern equipment and techniques, the risk of having a sick child in a particular family is determined or using the amneocetic method - the presence or absence of hereditary pathologies for a specific child. Sometimes families who doubt the relationship between parents (most often fathers) and children turn to the MGK. The MGK also provides answers to these questions with a certain degree of probability.

So, today we will try ourselves in the role of medical consultants, MHC patients and experts.

(The rules of the game are announced, instructions are given, and they are divided into groups).

Game "MGK".

Students, divided into groups, act in a certain role: “medical consultants”, “patients”, “experts”. The groups change roles. As a result, each student can try himself in different types of activities.

Sample problem options

(proposed by students)

№ 1. The first child in the family has phenylketonuria. Both parents are healthy (do not have phenotypic manifestations of the trait). What is the probability of having a healthy child in this family? What are the reasons for the birth of sick children from healthy parents?

№ 2* . There is a problem in the family: both spouses are dark-haired and brown-eyed, the spouses’ parents have the same characteristics, and the only child in the family is fair-haired and blue-eyed. The husband claims that this is impossible and accuses his wife of infidelity. Are your spouse's accusations justified?

№ 3. Two married couples doubt that they received their own daughters from the maternity hospital, because... Both girls were born at the same time, the birth was difficult, and the girls were only given identification tags the next day after birth. One pair of parents has blood groups I and III, and the second pair has blood groups II and III. Masha has blood type III, and Maya has blood type IV. Both parent couples claim that their daughter can only have blood type III and cannot have IV. How to solve a problem? DNA testing is a very expensive procedure and parents cannot afford it.

№ 4. A young married couple dreams of having a child, but the wife is afraid to have children, because... her brother died in early childhood from hemophilia. None of the relatives on the woman’s side suffered from hemophilia. Both spouses are also healthy. Is it possible for this couple to have children? What is the probability of having children with hemophilia in this family? How can you prevent the birth of a sick child?

Summarizing

At the end of the game, the results are summed up on the expert assessment sheets. The results are discussed.

Expert Evaluation Sheet

"patients"

Grade

F.I. "consultants"

Grade

Note: maximum score is 5.

Final lesson

Game "Relay Race"

Update: This form of lesson stimulates students’ cognitive activity, develops attention, the ability to formulate questions, and evaluate the work of other students.

Target : repetition and generalization of the material studied in the elective course.

Preparatory stage : advance task a week before the final lesson: prepare 1-2 questions on the topics studied.

Lesson plan

1. Teacher's opening speech. Assignment for the student.

   Relay race.

   Evaluation of questions and answers.

   Summarizing.

Relay rules

The named student answers the question asked, puts forward his hypotheses, arguments and asks his question to the next student, etc. until each student answers the question and asks his own. The number of questions depends on the number of students attending the course.

Evaluation of questions and answers. is given by the teacher and students in the evaluation sheets in the form of points placed opposite the student’s name: in the red column the score for the question, in the green for the answer, in the blue for originality and resourcefulness.

Based on the results of the analysis of the score sheets, winners are determined in three categories: “For the most interesting question”, “For the best answer”, “For originality and resourcefulness”.

Progress of the lesson

1. Introductory speech by the teacher.

Today we have the final lesson in the course “Biology in Problems”. During all this time, we learned to think, find solutions, ask questions, set goals, draw up assignments, observe, and tried to decide on the profile of study in high school. And today we sum up the results.

Laplace argued: “What we know is limited, and what we do not know is infinite.” We have solved many problems, but this is only a drop in the ocean of knowledge. The natural curiosity of man and the endless variety of life and the forms of its manifestation never cease to pose more and more new questions to man. And I suggest you conduct the final lesson in the form of a relay race. (Familiarization with the relay rules). You received an advanced task, which means you had time to search for material and create interesting assignments and questions. So, the relay starts.

Options for teacher questions

* To treat viral diseases, a medicine is used that contains the enzyme DNase. Is this medicine always effective? Why?

* What process does the reaction reflect: 2 H 2 O →H + + 4e + O 2. When does it happen?

Crossing short-tailed cats with each other always leads to the appearance of short-tailed and long-tailed kittens in the offspring. What are the genotypes of short-tailed cats?

The statistical regularity of the splitting law is splitting according to phenotype in a ratio of 3: 1. However, sometimes this ratio is violated. Name as many reasons for the statistical violation as possible.

2. Conducting the relay race

Students randomly ask questions prepared in advance to a specific student. The assessment of the proposed question is carried out by the answerer, and the answer by the questioner. The rest of the students and the teacher evaluate both the question and the answer. The assessment is carried out in three categories: correctness and knowledge of the question; completeness, logic and reasoning of the answer; originality and resourcefulness when answering.

3. Summing up

The counting commission selected from among the students determines the maximum total number of points for the question, answer, and originality. Based on the results, the winners are awarded “orders”: “For the most interesting question”, “For the best answer”, “For originality and resourcefulness”.

Evaluation paper

F.I. student

Question rating

Answer rating

Originality

and resourcefulness

Note: maximum score is 10.

Questionnaire options

Questionnaire 1.

Do you like the lessons in this course? Why?

What new have you learned?

What did you learn on the course?

Will you use the acquired knowledge in practice?

Did the course help you in choosing a training profile?

Questionnaire 2.

How did the course content meet your expectations?

(Underline whatever applicable):

Fully;

Partially (expected more, expected less);

Didn't match (expected more, expected less).

What methods and techniques were used most often by the teacher in the classroom? (Underline whatever applicable):

Conversation;

Lecture;

"Brainstorm";

Dispute;

Workshop;

Excursion;

Demonstration of experiments;

Project methodology.

Other (add)

Which of the methods used, in your opinion, were the most successful and interesting?

What did studying this course give you?

Did studying this course help you in choosing your future educational profile?

Your suggestions to the organizers of pre-profile training:

Reference materials

Number of chromosomes (2n)

Domestic dog

Chimpanzee

Homo sapiens

Drosophila

Horse roundworm

Potato

Soft wheat

Corn

fava beans

Peas

1. 1. 1. 1. 1. 1. Complete Domination

                  An object

                  Dominant trait

                  Recessive trait

                  Peas

                  yellow seeds

                  Green seeds

                  Smooth seeds

                  Wrinkled seeds

                  red corolla

                  White corolla

                  High growth

                  Dwarf stature

                  The fruit is white

                  Fruit yellow

                  Disc shape

                  spherical shape

                  Round form

                  Pear shape

                  Red fruit

                  yellow fruit

                  Early ripeness

                  Late ripening

                  Normal height

                  Gigantic growth

                  Drosophila

                  Red eyes

                  Cherry eyes

                  Gray body

                  Black body

                  Normal wings

                  Rudimentary wings

                  Guinea pig

                  Black wool

                  White wool

                  Brown wool

                  Long wool

                  Short hair

                  Shaggy wool

                  Smooth wool

                  Dark hair

                  Blonde hair

                  Red hair

                  Red hair

                  Normal pigmentation

                  Albinism

                  Brown eyes

                  Blue (gray) eyes

                  Big eyes

                  Small eyes

                  Thick lips

                  Thin lips

                  "Roman nose

                  Straight nose

                  Polydactyly

                  Normal hand structure

                  Short-fingered (brachydactyly)

                  Normal finger length

                  Freckles

                  No freckles

                  Short stature

                  Normal height

                  Normal hearing

                  Congenital deafness

                  Rh positive

                  Rh negativity

                  Incomplete dominance

An object

Signs of homozygotes

Signs

heterozygotes

Dominant

Recessive

Strawberries

Red fruit

White fruit

Pink fruit

Sweet pea

Red flower

White flower

Pink flower

Snapdragon

Red flower

White flower

Pink flower

Wide sheet

Narrow sheet

Middle sheet

Andalusian chickens

Black plumage

White plumage

Blue plumage

curly plumage

Smooth plumage

Wavy plumage

Cattle (cows)

Red wool

White wool

Roan wool

Black wool

White wool

Gray wool

Normal hemoglobin

Sickle cell anemia

Part of sickle-shaped red blood cells

Curly hair

Straight hair

Wavy hair

Sex-linked traits

An object

Dominant trait

Recessive trait

Normal blood clotting

Hemophilia

Normal color vision

Colorblindness

Normal development of sweat glands

Lack of sweat glands

Drosophila

Gray body color

Yellow body color

Red eye color

White eye color

Black coat color

Yellow coat color, tortoiseshell in heterozygotes

Traits determined by two interacting genes

An object

Determined by two non-allelic dominant genes (A+B)

Determined by two non-allelic recessive genes (a+b)

Determined by one dominant gene (A+b or a+B)

Gray (bay) color

Red suit

Black suit

Gray (agouti) color

White color

Black ( A+b) or white ( a+B)

red onion

White bulb

1. What is the difference between the concepts cell cycle and mitosis?

2. Scientists conducted research on mitosis: it turned out that in animals leading a nocturnal lifestyle, in most organs the maximum mitoses occur in the morning and the minimum at night. In diurnal animals, the maximum is observed in the evening, and the minimum during the day. Analyze this fact.

3. There is a phenomenon in which, after chromosome reproduction, cell division does not occur - endomitosis (Greek endo - inside). This leads to an increase in the number of chromosomes, sometimes tens of times. Endomitosis occurs, for example, in liver cells. What biological significance might this process have?

4. Why do you think scientists call the metaphase plate a kind of passport of the body?

5. Why can’t amitosis be considered a full-fledged method of cell reproduction, although this process occurs in connective tissue, in skin epithelial cells? In which cells do you think this method of division never occurs?