Methods of cytology material for preparing for the exam. Preparation for the exam: solving problems in cytology

The author of the article is D. A. Solovkov, Candidate of Biological Sciences

Types of tasks in cytology

The tasks in cytology that are found in the exam can be divided into seven main types. The first type is associated with determining the percentage of nucleotides in DNA and is most often found in part A of the exam. The second group includes computational tasks devoted to determining the number of amino acids in a protein, as well as the number of nucleotides and triplets in DNA or RNA. This type of problem can be found in both Part A and Part C.

Tasks in cytology types 3, 4 and 5 are devoted to working with a table of the genetic code, and also require knowledge of the transcription and translation processes from the applicant. Such tasks make up the majority of C5 questions in the exam.

Tasks of types 6 and 7 appeared in the USE relatively recently, and they can also be encountered by the applicant in part C. The sixth type is based on knowledge about changes in the genetic set of the cell during mitosis and meiosis, and the seventh type checks the student's assimilation of material on dissimilation in the eukaryotic cell .

Solutions to problems of all types are proposed below and examples for independent work are given. The appendix contains a table of the genetic code used in the solution.

Solving problems of the first type

Basic information:

• There are 4 types of nucleotides in DNA: A (adenine), T (thymine), G (guanine) and C (cytosine).
• In 1953, J. Watson and F. Crick discovered that the DNA molecule is a double helix.
• The chains are complementary to each other: opposite to adenine in one chain there is always thymine in the other and vice versa (A-T and T-A); opposite cytosine - guanine (C-G and G-C).
• In DNA, the amount of adenine and guanine is equal to the number of cytosine and thymine, as well as A=T and C=G (Chargaff's rule).

Problem: A DNA molecule contains adenine. Determine how many (in) this molecule contains other nucleotides.

Solution: the amount of adenine is equal to the amount of thymine, therefore, this molecule contains thymine. Guanine and cytosine account for . Because their numbers are equal, then C=G=.

Solving problems of the second type

Basic information:

• The amino acids required for protein synthesis are delivered to the ribosomes via tRNA. Each tRNA molecule carries only one amino acid.
• Information about the primary structure of a protein molecule is encrypted in the DNA molecule.
• Each amino acid is encoded by a sequence of three nucleotides. This sequence is called a triplet or codon.

Task: tRNA molecules participated in translation. Determine the number of amino acids that make up the resulting protein, as well as the number of triplets and nucleotides in the gene that codes for this protein.

Solution: if t-RNA was involved in the synthesis, then they transferred amino acids. Since one amino acid is encoded by one triplet, there will be triplets or nucleotides in the gene.

Solving problems of the third type

Basic information:

• Transcription is the process of synthesizing mRNA from a DNA template.
• Transcription is carried out according to the rule of complementarity.
• RNA contains uracil instead of thymine.

Task: a fragment of one of the DNA chains has the following structure: AAGGCTACGTTTG. Build i-RNA on it and determine the sequence of amino acids in a fragment of a protein molecule.

Solution: according to the rule of complementarity, we determine the mRNA fragment and divide it into triplets: UUC-CGA-UHC-AAU. According to the table of the genetic code, we determine the sequence of amino acids: phen-arg-cis-asn.

Solving problems of the fourth type

Basic information:

• An anticodon is a sequence of three nucleotides in tRNA that are complementary to the nucleotides of an mRNA codon. tRNA and mRNA contain the same nucleotides.
• The mRNA molecule is synthesized on DNA according to the rule of complementarity.
• DNA contains thymine instead of uracil.

Task: i-RNA fragment has the following structure: GAUGAGUATSUUTCAAA. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write down the fragment of the DNA molecule on which this mRNA was synthesized.

Solution: break the mRNA into GAU-GAG-UAC-UUC-AAA triplets and determine the amino acid sequence using the genetic code table: asp-glu-thir-phen-lys. This fragment contains triplets, so t-RNA will participate in the synthesis. Their anticodons are determined according to the rule of complementarity: CUA, CUC, AUG, AAG, UUU. Also, according to the rule of complementarity, we determine the DNA fragment (by i-RNA !!!): TSTATSTSATGAAGTTT.

Solving problems of the fifth type

Basic information:

• The tRNA molecule is synthesized on DNA according to the rule of complementarity.
• Remember that RNA contains uracil instead of thymine.
• An anticodon is a sequence of three nucleotides that are complementary to the nucleotides of a codon in mRNA. tRNA and mRNA contain the same nucleotides.

Task: a DNA fragment has the following nucleotide sequence TTAGCCGATCCG. Set the nucleotide sequence of the tRNA that is synthesized on this fragment and the amino acid that this tRNA will carry if the third triplet corresponds to the tRNA anticodon. To solve the problem, use the table of the genetic code.

Solution: we determine the composition of the t-RNA molecule: AAUCGGCUAGGC and find the third triplet - this is CUA. This anticodon is complementary to the i-RNA triplet - GAU. It codes for the amino acid asp, which is carried by this tRNA.

Solving problems of the sixth type

Basic information:

• The two main types of cell division are mitosis and meiosis.
• Changes in the genetic makeup of a cell during mitosis and meiosis.

Task: in an animal cell, the diploid set of chromosomes is equal to. Determine the number of DNA molecules before mitosis, after mitosis, after the first and second divisions of meiosis.

Solution: By condition, . Genetic set:

Solving problems of the seventh type

Basic information:

• What is metabolism, dissimilation and assimilation.
• Dissimilation in aerobic and anaerobic organisms, its features.
• How many stages in dissimilation, where they go, what chemical reactions take place during each stage.

Task: glucose molecules entered into dissimilation. Determine the amount of ATP after glycolysis, after the energy stage and the total effect of dissimilation.

Solution: write the glycolysis equation: \u003d 2PVC + 4H + 2ATP. Since PVC and 2ATP molecules are formed from one glucose molecule, therefore, 20 ATP are synthesized. After the energy stage of dissimilation, ATP molecules are formed (during the breakdown of a glucose molecule), therefore, ATP is synthesized. The total effect of dissimilation is equal to ATP.

Examples of tasks for independent solution

1. T=, G=C= by .
2. amino acids, triplets, nucleotides.
3. triplet, amino acids, t-RNA molecules.
4. i-RNA: CCG-AGA-UCG-AAG. Amino acid sequence: pro-arg-ser-lys.
5. DNA fragment: CGATTACAAGAAATG. tRNA anticodons: CGA, UUA, CAA, GAA, AUG. Amino acid sequence: ala-asn-val-ley-tir.
6. t-RNA: UCG-GCU-GAA-CHG. The anticodon is GAA, the codon of i-RNA is CUU, the transferred amino acid is leu.
7. . Genetic set:
8. Since PVC and 2ATP molecules are formed from one glucose molecule, therefore, ATP is synthesized. After the energy stage of dissimilation, ATP molecules are formed (during the breakdown of a glucose molecule), therefore, ATP is synthesized. The total effect of dissimilation is equal to ATP.
9. PVC molecules entered the Krebs cycle, therefore, glucose molecules broke up. The amount of ATP after glycolysis - molecules, after the energy stage - molecules, the total effect of dissimilation of ATP molecules.

So, this article lists the main types of tasks in cytology that an applicant can meet in the exam in biology. We hope that the variants of tasks and their solution will be useful to everyone in preparing for the exam. Good luck!

Cell theory, its main provisions, role in the formation of the modern natural-science picture of the world. Development of knowledge about the cell. The cellular structure of organisms, the similarity of the structure of the cells of all organisms - the basis of the unity of the organic world, evidence of the relationship of living nature.

A cell is a unit of structure, life activity, growth and development of organisms. variety of cells. Comparative characteristics of cells of plants, animals, bacteria, fungi.

The structure of pro- and eukaryotic cells. The relationship of the structure and functions of the parts and organelles of the cell is the basis of its integrity. Metabolism: energy and plastic metabolism, their relationship. Enzymes, their chemical nature, role in metabolism. Stages of energy metabolism. Fermentation and respiration. Photosynthesis, its significance, cosmic role. Phases of photosynthesis. Light and dark reactions of photosynthesis, their relationship. Chemosynthesis.

Biosynthesis of proteins and nucleic acids. Matrix nature of biosynthetic reactions. Genes, genetic code and its properties. Chromosomes, their structure (shape and size) and functions. The number of chromosomes and their species constancy. Determination of the set of chromosomes in somatic and germ cells. Cell life cycle: interphase and mitosis. Mitosis is the division of somatic cells. Meiosis. Phases of mitosis and meiosis. The development of germ cells in plants and animals. Similarities and differences between mitosis and meiosis, their significance. Cell division is the basis for the growth, development and reproduction of organisms.

SOLVING PROBLEMS IN CYTOLOGY

Teaching aid

New Adelakovo 2014

Compiled by: biology teacher L.I. Denisova

Solving problems in cytology. For students in grades 9-11 / GBOU secondary school v. Novoe Adelakovo "; comp. L.I.Denisova. – New Adelakovo, 2014.

The teaching aid has been compiled for the purpose of use by biology teachers and students in preparation for the final certification, as well as for biology olympiads. The manual contains brief theoretical material on each type of problem and examples of problem solving. Designed for students in grades 9-11 of educational institutions and biology teachers.

Introduction……………………………………………………………..

2.1 Tasks related to determining the percentage of nucleotides in DNA

2.2. R

2.3. Tasks for the construction of an i-RNA molecule, t-RNA anticodons and amino acid sequences in a polypeptide chain. Work

Literature.

Introduction.

The solution of problems in cytology is included in the KIM in biology in the Unified State Examination (39 task). The completion of this task provides for a detailed answer and is aimed at testing skills.

• apply knowledge in a new situation;
• establish causal relationships;
• analyze, systematize and integrate knowledge;
• summarize and formulate conclusions.

According to the results of the analysis of the results of the USE in biology, the following are among the poorly formed knowledge and skills of the participants in the exam:

1. determination of the chromosome set of cells in plant development cycles;
2. determination of the number of chromosomes and DNA in cells in different phases of mitosis and meiosis;
3. explanation and justification of the result.

When studying biology at a basic level, there is not enough time to practice solving problems in cytology. After passing through the relevant topics, without constant repetition of the practical development of problem solving, the skills are quickly forgotten. Students can have this manual always at hand in order to remember the course of solving typical problems. Especially in rural areas, not everyone has free access to Internet resources.

Tasks in cytology that are found in the exam can be divided into several main types. This manual offers solutions to problems of various types and provides examples for independent work. The appendix contains a table of the genetic code used in the solution.

This teaching aid has been compiled to help biology teachers studying in grades 9-11 of secondary schools.

1.1 Key points to remember when solving problems in cytology.

1. Each amino acid is delivered to the ribosomes of one tRNA, therefore,the number of amino acids in a protein is equal to the number of tRNA moleculesinvolved in protein synthesis;
2. each amino acid is encoded by three nucleotides (one triplet, or codon), so the number of coding nucleotides is always three times greater, andthe number of triplets (codons) is equal to the number of amino acids in the protein;
3. each tRNA has an anticodon that is complementary to an mRNA codon, sonumber of anticodons, and therefore in general tRNA molecules equal to the number of mRNA codons;
4. mRNA is complementary to one of the DNA strands, sothe number of mRNA nucleotides is equal to the number of DNA nucleotides. The number of triplets, of course, will also be the same.

When solving a number of problems in this section, it is necessary to use the table of the genetic code. The rules for using the table are usually indicated in the task, but it is better to learn this in advance. To determine the amino acid encoded by a particular triplet, you must perform the following steps:

1. the first nucleotide of the triplet is found in the left vertical row,
2. the second - in the upper horizontal,
3. the third is in the right vertical row.
4. the amino acid corresponding to the triplet is located at the intersection point of the imaginary lines, coming from nucleotides.

1.2. Written formulation of problem solving.

Answer:

An important point in solving tasks is an explanation of the actions performed, especially if the task says so: "Explain the answer." The presence of explanations allows the verifier to conclude that the student understands the topic, and their absence can lead to the loss of a very important score. Task 39 is estimated at three points, which are awarded in case of a completely correct solution. Therefore, starting the task, first of all, it is necessary to identify all the questions. The number of responses must match.

Chapter 2. Types of tasks in cytology.

2.1 Tasks related to determining the percentage of nucleotides in DNA

Even before the discovery of Watson and Crick, in 1950, the Australian biochemist Edwin Chargaff established thatthat in the DNA of any organism the number of adenyl nucleotides is equal to the number of thymidyl, and the number of guanyl nucleotides is equal to the number of cytosyl nucleotides (A=T, G=C), or the total number of purine nitrogenous bases is equal to the total number of pyrimidine nitrogenous bases (A+G=C+T ).These patterns are called "Chargaff's rules".

The fact is that when a double helix is ​​formed, the nitrogenous base of adenine in one chain is always opposite to the nitrogenous base of adenine in the other chain, and opposite to guanine is cytosine, that is, the DNA chains seem to complement each other. And these paired nucleotides are complementary to each other (from Latin complementum - addition).

Why is this principle followed? To answer this question, we need to remember the chemical nature of nitrogenous heterocyclic bases. Adenine and guanine belong to purines, and cytosine and thymine belong to pyrimidines, that is, bonds between nitrogenous bases of the same nature are not established. In addition, complementary bases correspond geometrically to each other, i.e. in size and shape. Thus, the complementarity of nucleotides is the chemical and geometric correspondence of the structures of their molecules to each other.

Nitrogenous bases contain strongly electronegative oxygen and nitrogen atoms, which carry a partial negative charge, as well as hydrogen atoms, on which a partial positive charge arises. Due to these partial charges, hydrogen bonds arise between the nitrogenous bases of the antiparallel sequences of the DNA molecule.

Task. In a DNA molecule, there are 23% of adenyl nucleotides of the total number of nucleotides. Determine the amount of thymidyl and cytosyl nucleotides.

Answer: T=23%; C=27%

Task Given a DNA molecule with a relative molecular weight of 69 thousand, of which 8625 are adenyl nucleotides. The relative molecular weight of one nucleotide is on average 345. How many nucleotides are there individually in this DNA? What is the length of its molecule?

Answer : A=T=25; G=C=75; 34 nm.

2.2. R computational tasks devoted to determining the number of amino acids in a protein, as well as the number of nucleotides and triplets in DNA or RNA.

• The amino acids required for protein synthesis are delivered to the ribosomes via tRNA. Each tRNA molecule carries only one amino acid.
• Information about the primary structure of a protein molecule is encrypted in the DNA molecule.
• Each amino acid is encoded by a sequence of three nucleotides. This sequence is called a triplet or codon.

Task: 30 t-RNA molecules participated in translation. Determine the number of amino acids that make up the resulting protein, as well as the number of triplets and nucleotides in the gene that codes for this protein.

Answer : Number of a / c - 30. Number of triplets - 30. Number of nucleotides - 90.

Task: According to some scientists, the total length of all DNA molecules in the nucleus of one human germ cell is about 102 cm. How many base pairs are there in the DNA of one cell (1 nm = 10–6 mm)?

Answer: 3x109 pairs.

2.3. Tasks for the construction of an i-RNA molecule, t-RNA anticodons and amino acid sequences in a polypeptide chain. Workwith a table of the genetic code.

Task: RNA contains uracil instead of thymine. Protein biosynthesis involved t-RNAs with anticodons: UUA, GHC, CHC, AUU, CGU. Determine the nucleotide sequence of the section of each chain of the DNA molecule that carries information about the synthesized polypeptide, and the number of nucleotides containing adenine, guanine, thymine, cytosine in a double-stranded DNA molecule

Task: a fragment of one of the DNA chains has the following structure: AAGGTSTACGTTG. Build i-RNA on it and determine the sequence of amino acids in a fragment of a protein molecule.

Answer: fen-arg-cis-asn.

Task: Ribosomes from different cells, the entire set of amino acids and the same molecules of i-RNA and t-RNA were placed in a test tube, creating all the conditions for protein synthesis. Why in a test tube one type of protein will be synthesized on different ribosomes.

Answer: On the same mRNA, the same protein is synthesized, since the information is the same.

• Task: i-RNA fragment has the following structure: GAUGAGUATSUUTCAAA. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write down the fragment of the DNA molecule on which this mRNA was synthesized.

Answer : Amino acid sequence - : asp-glu-thir-phen-lys. tRNA anticodons - CUA, CUC, AUG, AAG, UUU. DNA fragment - CTACTCATGAAGTTT

2.4. Tasks for determining the number of DNA molecules and chromosomes in the process of mitosis and meiosis.

Mitosis - the main method of division of eukaryotic cells, in which doubling first occurs, and then a uniform distribution between the daughter cells of the hereditary material.

Mitosis is a continuous process in which there are four phases: prophase, metaphase, anaphase, and telophase. Before mitosis, the cell prepares for division, or interphase. The period of cell preparation for mitosis and mitosis itself together make upmitotic cycle.

Interphase consists of three periods: presynthetic, or postmitotic, - G 1 , synthetic - S, postsynthetic, or premitotic - G 2 .

Presynthetic period(2 n 2 c , where n - the number of chromosomes, with - the number of DNA molecules) - cell growth, activation of biological synthesis processes, preparation for the next period.

Synthetic period(2n4s ) - DNA replication.

ATTENTION!

After doubling, the chromosome set remains diploid (2n) because the sister chromatids remain connected at the centromere.

Postsynthetic period(2n4s ) - preparation of the cell for mitosis, synthesis and accumulation of proteins and energy for the upcoming division, an increase in the number of organelles, doubling of centrioles.

Prophase (2 n 4 s ) - dismantling of nuclear membranes, divergence of centrioles to different poles of the cell, formation of fission spindle threads, "disappearance" of nucleoli, condensation of two-chromatid chromosomes.

Metaphase (2 n 4 s ) - alignment of the most condensed two-chromatid chromosomes in the equatorial plane of the cell (metaphase plate), attachment of the spindle fibers with one end to the centrioles, the other - to the centromeres of the chromosomes.

Anaphase (4 n 4 s ) - the division of two-chromatid chromosomes into chromatids and the divergence of these sister chromatids to opposite poles of the cell (whilechromatids become independent single chromatid chromosomes).

Telophase (2 n 2 s in each daughter cell) - decondensation of chromosomes, the formation of nuclear membranes around each group of chromosomes, the disintegration of the fission spindle threads, the appearance of the nucleolus, the division of the cytoplasm (cytotomy). Cytotomy in animal cells occurs due to the fission furrow, in plant cells - due to the cell plate.

Meiosis - this is a special way of dividing eukaryotic cells, as a result of which the transition of cells from a diploid state to a haploid one occurs. Meiosis consists of two consecutive divisions preceded by a single DNA replication.

First meiotic division (meiosis 1)called reduction, because it is during this division that the number of chromosomes is halved: from one diploid cell (2 n 4 s ) form two haploid (1 n 2 c ).

Interphase 1 (at the beginning - 2 n 2 c, at the end - 2 n 4 c ) - synthesis and accumulation of substances and energy necessary for the implementation of both divisions, an increase in cell size and the number of organelles, doubling of centrioles, DNA replication, which ends in prophase 1.

Prophase 1 (2 n 4 s ) - dismantling of nuclear membranes, divergence of centrioles to different poles of the cell, formation of fission spindle threads, "disappearance" of nucleoli, condensation of two-chromatid chromosomes, conjugation of homologous chromosomes and crossing over.

Metaphase 1 (2 n 4 s ) - alignment of bivalents in the equatorial plane of the cell, attachment of the fission spindle threads at one end to the centrioles, the other - to the centromeres of the chromosomes.

Anaphase 1 (2 n 4 s ) - random independentdivergence of two-chromatid chromosomesto opposite poles of the cell (from each pair of homologous chromosomes, one chromosome goes to one pole, the other to the other), recombination of chromosomes.

Telophase 1 (1 n 2 s in each cell) - the formation of nuclear membranes around groups of two-chromatid chromosomes, the division of the cytoplasm. In many plants, a cell from anaphase 1 immediately transitions to prophase 2.

Second meiotic division (meiosis 2) is called equational.

Interphase 2, or interkinesis (1n 2c ), is a short break between the first and second meiotic divisions during which DNA replication does not occur. characteristic of animal cells.

Prophase 2 (1 n 2 s ) - dismantling of nuclear membranes, divergence of centrioles to different poles of the cell, the formation of fission spindle filaments.

Metaphase 2 (1 n 2 s ) - alignment of two-chromatid chromosomes in the equatorial plane of the cell (metaphase plate), attachment of the spindle fibers with one end to the centrioles, the other - to the centromeres of the chromosomes; 2 block of oogenesis in humans.

Anaphase 2 (2 n 2 s ) - the division of two-chromatid chromosomes into chromatids and the divergence of these sister chromatids to opposite poles of the cell (in this case, the chromatids become independent single-chromatid chromosomes), recombination of chromosomes.

Telophase 2 (1 n 1 s in each cell) - decondensation of chromosomes, the formation of nuclear membranes around each group of chromosomes, the disintegration of the fission spindle threads, the appearance of the nucleolus, the division of the cytoplasm (cytotomy) with the formation of four haploid cells as a result.

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

Answer: Interphase contains 60 chromosomes and 120 DNA molecules. After meiosis I - 30 chromosomes and 60 DNA molecules.

Task: The chromosome set of wheat somatic cells is 28. Determine the chromosome set and the number of DNA molecules in the nucleus (cell) of the ovule before meiosis I and meiosis II. Explain the results in each case.

Answer: Before meiosis I = 28 chromosomes, 56 DNA molecules. Before meiosis II = 14 chromosomes, 28 DNA molecules

Task: It is known that somatic cells of cabbage contain 18 chromosomes. Determine the chromosome set and the number of DNA molecules in one of the cells of the ovule before the start of meiosis, in anaphase of meiosis I and in anaphase of meiosis II. Explain what processes take place during these periods and how they affect the change in the number of DNA and chromosomes.

Answer: Before the start of meiosis - 18 chromosomes, 36 DNA molecules. Anaphase I of meiosis has 18 chromosomes and 36 DNA molecules. In anaphase, there are 18 chromosomes, 18 DNA molecules.

2.5. Tasks to determine the length of a single section of DNA or the number of nucleotides in it.

According to the model of the American biochemist J. Watson and the English physicist F. Crick, DNA molecules are two right-handed polynucleotide chains around a common axis, or a double helix. There are approximately 10 nucleotide residues per turn of the helix. The strands in this double helix are anti-parallel, that is, they point in opposite directions, so that the 3" end of one strand is opposite the 5" end of the other. The sizes of DNA molecules are usually expressed by the number of nucleotides that form them. These sizes vary from several thousand base pairs in bacterial plasmids and some viruses to many hundreds of thousands of base pairs in higher organisms.

Linear length of one nucleotide in nucleic acid

L n \u003d 0.34 nm \u003d 3.4 angstroms

Average molecular weight of one nucleotide

Mr n = 345 a.m.u. (Yes)

Task. The contour length of a bacteriophage DNA molecule is 17x10" 6 m. After exposure to mutagens, the length turned out to be 13.6x10-6 m. Determine how many pairs of nitrogenous bases have dropped out as a result of a mutation, if it is known that the distance between adjacent nucleotides is 34x10 11 m.

Answer: 10 thousand base pairs.

Chapter 4. Examples of tasks for independent solution.

1. The DNA molecule contains 31% adenine. Determine how many (in%) this molecule contains other nucleotides.
2. Translation involved 50 t-RNA molecules. Determine the number of amino acids that make up the resulting protein, as well as the number of triplets and nucleotides in the gene that codes for this protein.
3. A DNA fragment consists of 72 nucleotides. Determine the number of triplets and nucleotides in mRNA, as well as the number of amino acids that make up the resulting protein.
4. A fragment of one of the DNA chains has the following structure: GGCTCTAGCTTTC. Build i-RNA on it and determine the sequence of amino acids in a fragment of a protein molecule (for this, use the table of the genetic code).
5. The i-RNA fragment has the following structure: ГЦУАУГУУУУУУУКАЦ. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write down the fragment of the DNA molecule on which this mRNA was synthesized (for this, use the table of the genetic code).
6. The DNA fragment has the following nucleotide sequence AGCCGACCTTGCCC. Set the nucleotide sequence of the tRNA that is synthesized on this fragment and the amino acid that this tRNA will carry if the third triplet corresponds to the tRNA anticodon. To solve the problem, use the table of the genetic code.
7. In an animal cell, the diploid set of chromosomes is 20. Determine the number of DNA molecules before mitosis, after mitosis, after the first and second divisions of meiosis.
8. 15 glucose molecules entered into dissimilation. Determine the amount of ATP after glycolysis, after the energy stage and the total effect of dissimilation.
9. 6 PVC molecules entered the Krebs cycle. Determine the amount of ATP after the energy stage, the total effect of dissimilation and the number of glucose molecules that have entered into dissimilation.

Answers:

1. T=31%, G=C=19% each.
2. 50 amino acids, 50 triplets, 150 nucleotides.
3. 24 triplets, 24 amino acids, 24 tRNA molecules.
4. i-RNA: CCG-AGA-UCG-AAG. Amino acid sequence: pro-arg-ser-lys.
5. DNA fragment: CGATTACAAGAAATG. tRNA anticodons: CGA, UUA, CAA, GAA, AUG. Amino acid sequence: ala-asn-val-ley-tir.
6. t-RNA: UCG-GCU-GAA-CHG. Anticodon GAA, codon i-RNA - CUU, portable amino acid - leu.
7. 2n=20. Genetic set:

1. before mitosis, 40 DNA molecules;
2. after mitosis, 20 DNA molecules;
3. after the first division of meiosis, 20 DNA molecules;
4. after the second division of meiosis, 10 DNA molecules.

1. Since 2 molecules of PVC and 2ATP are formed from one glucose molecule, therefore, 30 ATP are synthesized. After the energy stage of dissimilation, 36 ATP molecules are formed (during the breakdown of 1 glucose molecule), therefore, 540 ATP are synthesized. The total effect of dissimilation is 540+30=570 ATP.
2. 6 PVC molecules entered the Krebs cycle, therefore, 3 glucose molecules decomposed. The amount of ATP after glycolysis - 6 molecules, after the energy stage - 108 molecules, the total effect of dissimilation of 114 ATP molecules.

Literature:

Control tasks in genetics and guidelines for their implementation. [comp. L.I. Lushina, S.V. Zalyashchev, A.A. Semenov, O.N. Noskova]. - Samara: SGPU, 2007, 142p.

http://ege-study.ru

http://licey.net

http://reshuege.ru

http://www.fipi.ru

Annex I Genetic code (i-RNA)

In the lesson, we will learn the history of the emergence of cytology, recall the concept of a cell, and consider what contribution various scientists have made to the development of cytology.

All living creatures, with the exception of vi-ru-owls, consist of cells. But for the scientists of the past, the cellular structure of living or-ga-niz-ms was not as obvious as it was for you and me. Science, studying the cell-ku, cytology, sfor-mi-ro-va-las only to the middle of the 19th century. Without knowing where life comes from, what is its small-tea-shey unity, up to the Middle-ne-ve-ko-vya there were theories about, for example, that frogs pro-is-ho-dy from dirt, and mice are born in dirty underwear (Fig. 2).

Rice. 2. Theories of the Middle Ages ()

“Dirty linen of middle-world-science” was the first “time-to-sew” in 1665. bert Hooke (Fig. 3).

Rice. 3. Robert Hooke ()

For the first time, he looked at and described the shells of the growing cells. And already in 1674, his Dutch colleague An-to-ni van Lee-wen-hoek (Fig. 4) for the first time looked under the self-del-miik -ro-sko-pom of some simple-shih and separate animal cells, such as erit-ro-qi-you and sper-ma-to-zo-i -dy.

Rice. 4. Anthony van Leeuwenhoek ()

Is-sle-before-va-nia Le-wen-gu-ka-ka-za-lis-with-time-men-ni-kam on-so-fan-ta-sti-che-ski-mi that in 1676 the year of the London-Don Ko-ro-left-society, where he from-sy-lal re-zul-ta-you of his research-to-va-ny, very strong in them for-with-me-wa-moose. The existence of one-but-kle-toch-nyh or-ga-niz-ms and blood cells, for example, does not fit in the framework of that old science.

In order to comprehend the results of the work of the Dutch scientist, it took several centuries. Only to the middle of the XIX century. German scientist Theodor Schwann, based on the work of his colleague Ma-tti-a-sa Schlei-de-na (Fig. 5 ), sfor-mu-li-ro-shaft of the basis of the new-lo-same-tion of the exact theory, which we still use to this day.

Rice. 5. Theodor Schwann and Matthias Schleiden ()

Schwann do-ka-zal that the cells of races and animals have a common principle of structure, because they form the same on-to-you spo-so-bom; all cells are sa-mo-sto-i-tel-na, and any or-ga-nism is a combination of life-not-de-i-tel-no-sti from del-ny groups of cells (Fig. 6).

Rice. 6. Red blood cells, cell division, DNA molecule ()

Further studies of scientific positions-whether-whether sfor-mu-whether-ro-vat basics-new-states-of-the-time -noy kle-toch-noy theory:

1. The cage is a universal structural unit of life.
2. Cells are multiplied by de-le-tion (cell from cell).
3. Cells are stored, re-re-ra-ba-you-va-yut, re-a-li-zu-yut and re-re-da-yut on-the-sequence-in-form-ma -tion.
4. A cell is a sa-mo-hundred-I-tel-naya bio-si-ste-ma, from-ra-zha-yu-shchaya opre-de-len-ny structural level of or- ga-ni-za-tion living ma-te-rii.
5. Many-clear-accurate or-ga-bottom-we are a complex of inter-and-mo-acting-stu-ing systems of various cells, providing chi-va-yu-shchy or-ga-bottom-mu growth, development, exchange of substances and energy.
6. The cells of all or-ga-niz-mov are similar to each other in terms of structure, chi-mi-che-sko-mu co-hundred and functions.

Cells through-you-tea-but once-but-about-times. They can differ in structure, form and function (Fig. 7).

Rice. 7. Diversity of cells ()

Among them there are free-but-living cells, some of them behave like individuals of populations and species, like self-sto-I-tel-nye- ha-bottom-we. Their life-not-de-I-tel-ness depends not only on how ra-bo-ta-yut inside-ri-kle-precise structures-tu-ry, or-ga -but-and-dy. They themselves need to get their own food, move around in the environment, multiply, that is, act to be like small, but quite self-worthy individuals. There are a lot of such free-to-lu-bi-out one-but-kle-toch-nyh. They enter all the kingdoms of cellular living nature and on-se-la-yut all the environments of life on our planet. In a many-cle-precise or-ga-bottom-me, a cell is-la-is-a part of it, tissues and or-ha are formed from cells -us.

The sizes of cells can be very different - from one de-xia-that mik-ro-on and up to 15 san-ti-meters - this is the size of the straw-y-sa egg , representing one cell-ku, and the weight of this cell-ki is half-to-ra ki-lo-gram-ma. And this is not the limit: di-no-zav-ditch eggs, for example, could reach a length of as much as 45 san-ti-meters (Fig. 8) .

Rice. 8. Dinosaur egg ()

Usually, in many-cell-accurate or-ga-niz-movs, different cells perform different functions. Cells, similar in structure, arranged side by side, united by inter-cellular substance and pre-sign -chennye for performing certain functions in the or-ga-bottom, form tissues (Fig. 9).

Rice. 9. Tissue formation ()

The life of a many-cle-toch-no-go or-ga-niz-ma for-ve-sits on how-so-co-wife-but-ra-bo-ta-yut cells enter dya-schee in his composition. Therefore, the cells do not con-ku-ri-ru-yut among themselves, on-against, co-operation and special-a-li-za-tion of their functions pos-in-la-et or-ga-bottom-mu you-live in those si-tu-a-qi-yah, in some one-night-cells you don’t-live-va- ut. In complex many-kle-toch-nyh or-ga-niz-mov - races, animals and people-lo-ve-ka - cells-ki or-ga-ni- zo-va-ny in fabrics, fabrics - in org-ga-ny, org-ga-ny - in si-ste-we org-new. And each of these systems works to provide the existence of the whole or-ga-niz-mu.

Despite all the different shapes and sizes, cells of different types are similar to each other. Such processes as respiration, bio-synthesis, metabolism, go on in cells regardless of whether they are one -no-kle-toch-ny-mi or-ga-niz-ma-mi or are part of many-kle-toch-no-th-essence. Each cell swallows food, draws energy from it, creatures, under-der-zhi-va-et in-hundred-yan-stvo of its own chi-mi-che-so-hundred-va and re-pro-from-in-dit itself, that is, it carries out all the pro-cesses, from someone it depends on her life.

All this pos-vo-la-et ras-smat-ri-vat the cell as a special unit of living ma-ter-rii, as an element-men-tar-living system ( Fig. 10).

Rice. 10. Schematic drawing of a cell ()

All living creatures, from in-fu-zo-rii to an elephant or a whale, sa-mo-go large-no-go on this day-to-day-to-pi-ta-yu- more, so-hundred-yat from cells. The only difference is that the in-fu-zo-rii are sa-mo-hundred-I-tel-nye bio-si-ste-we, consisting of a hundred-I-s from one cell, and the cells of the whale are or-ga-ni-zo-va-ny and vza-and-mo-connected-for-us as parts of a big-sho-go 190-ton-no-th whole. The composition of the whole or-ga-niz-ma is for-vi-sit on how its parts function, that is, cells.

Bibliography

1. Mamontov S.G., Zakharov V.B., Agafonova I.B., Sonin N.I. Biology. General patterns. - Bustard, 2009.
2. Ponomareva I.N., Kornilova O.A., Chernova N.M. Fundamentals of General Biology. Grade 9: A textbook for students in grade 9 educational institutions / Ed. prof. I.N. Ponomareva. - 2nd ed., revised. - M.: Ventana-Graf, 2005
3. Pasechnik V.V., Kamensky A.A., Kriksunov E.A. Biology. An Introduction to General Biology and Ecology: A 9th Grade Textbook, 3rd ed., stereotype. - M.: Bustard, 2002.

1. Krugosvet.ru ().
2. Uznaem-kak.ru ().
3. Mewo.ru ().

Homework

1. What does cytology study?
2. What are the main provisions of the cell theory?
3. How are cells different?

Dear readers! If you choose the USE as your final or entrance examination in biology, then you need to know and understand the requirements for passing this exam, the nature of the questions and tasks found in the examination papers. To help applicants, the EKSMO publishing house will publish the book “Biology. Collection of tasks for preparing for the exam. This book is a training manual, which is why the material included in it exceeds the school level of requirements. However, for those high school students who decide to enter higher educational institutions at faculties where biology is taken, this approach will be useful.

In our newspaper, we publish only the tasks of part C for each section. They are completely updated both in content and in the structure of presentation. Since this manual is focused on the exams of the 2009/2010 academic year, we decided to give options for Part C tasks in a much larger volume than was done in previous years.

You are offered approximate options for questions and tasks of different levels of complexity with a different number of elements of the correct answer. This is done so that on the exam you have a sufficiently large selection of possible correct answers to a specific question. In addition, the questions and tasks of Part C are structured as follows: one question and the elements of the correct answer to it are given, and then variants of this question are offered for independent reflection. The answers to these options should be obtained by you yourself, applying both the knowledge gained from studying the material and the knowledge gained from reading the answers to the main question. All questions must be answered in writing.

A significant part of the tasks of part C are tasks in drawings. Similar to them were already in the examination papers of 2008. In this manual, their set is somewhat expanded.

We hope that this textbook will help high school students not only prepare for exams, but also provide an opportunity for those who wish to learn the basics of biology in the remaining two years of study in grades 10-11.

General Biology (Part C)

The tasks of this part are divided into sections: cytology, genetics, evolutionary theory, ecology. Each section offers tasks for all levels of the exam. Such a construction of the general biological part of the manual will allow you to more fully and systematically prepare for the exam, because. Part C includes, in a generalized form, almost all of the material in Parts A and B.

Tasks of group C1 (advanced level)

All tasks of group C must be answered in writing with explanations.

Questions on cytology

The answer to this question should be short but precise. The words “levels of organization” and “scientific foundations” are the main ones in this matter. The level of organization is the way and form of existence of living systems. For example, the cellular level of organization includes cells. Therefore, it is necessary to find out what is common, which made it possible to distinguish the levels of organization. Such a common feature is the systematic organization of living bodies and their gradual complication (hierarchy).

Elements of the correct answer

The scientific grounds for dividing living systems into levels are the following provisions.

1. Living systems become more complex as they develop: cell - tissue - organism - population - species, etc.

2. Each more highly organized living system includes the previous systems. Tissues are made up of cells, organs are made up of tissues, organisms are made up of organs, and so on.

Answer the following questions yourself

What common properties do all levels of life organization have?

What are the similarities and differences between the cellular and population levels of life?

Prove that all the properties of living systems are manifested at the cellular level.

Elements of the correct answer

1. It is possible to apply influences to the model that are not applicable to living bodies.

2. Modeling allows you to change any characteristics of the object.

Answer yourself

How would you explain the statement of I.P. Pavlova “Observation collects what nature offers it, while experience takes from nature what it wants”?

Give two examples of the use of the experimental method in cytology.

What research methods can be used to separate different cellular structures?

Elements of the correct answer

1. The polarity of a water molecule determines its ability to dissolve other hydrophilic substances.

2. The ability of water molecules to form and break hydrogen bonds between them provides water with heat capacity and thermal conductivity, the transition from one state of aggregation to others.

3. The small size of the molecules ensures their ability to penetrate between the molecules of other substances.

Answer yourself

What will happen to the cell if the concentration of salts in it is higher than outside the cell?

Why don't cells shrink and burst from swelling in physiological saline?

Elements of the correct answer

1. Scientists have found that a protein molecule has primary, secondary, tertiary and quaternary structures.

2. Scientists have found that a protein molecule consists of many different amino acids linked by peptide bonds.

3. Scientists have established the sequence of amino acid residues in the ribonuclease molecule, i.e. its primary structure.

Answer yourself

What chemical bonds are involved in the formation of a protein molecule?

What factors can lead to protein denaturation?

What are the features of the structure and functions of enzymes?

In what processes are the protective functions of proteins manifested?

Elements of the correct answer

1. These organic compounds perform a building (structural) function.

2. These organic compounds perform an energy function.

Answer yourself

Why is cellulose-rich food prescribed to normalize bowel function?

What is the building function of carbohydrates?

Elements of the correct answer

1. DNA is built on the principle of a double helix in accordance with the rule of complementarity.

2. DNA consists of repeating elements - 4 types of nucleotides. Different sequence of nucleotides encodes different information.

3. The DNA molecule is capable of self-reproduction, and therefore, of copying information and its transmission.

Answer yourself

What facts prove the individuality of the DNA of an individual?

What does the concept of "universality of the genetic code" mean; what facts confirm this universality?

What is the scientific merit of D. Watson and F. Crick?

Elements of the correct answer

1. Differences in the names of DNA and RNA are explained by the composition of their nucleotides: in DNA nucleotides, the carbohydrate is deoxyribose, and in RNA, ribose.

2. Differences in the names of RNA types (informational, transport, ribosomal) are associated with the functions they perform.

Answer yourself

What two conditions must be constant for the bonds between two complementary DNA strands not to break spontaneously?

How do DNA and RNA differ in structure?

What other compounds contain nucleotides, and what do you know about them?

Elements of the correct answer

1. Cell theory established the structural and functional unit of the living.

2. The cell theory established the unit of reproduction and development of the living.

3. Cell theory confirmed the common structure and origin of living systems.

Answer yourself

Why, despite the obvious differences in the structure and functions of the cells of different tissues, they talk about the unity of the cellular structure of the living?

What are the main discoveries in biology that made it possible to formulate the cell theory.

Elements of the correct answer

1. Substances enter the cell by diffusion.

2. Substances enter the cell due to active transport.

3. Substances enter the cell by pinocytosis and phagocytosis.

Answer yourself

What is the difference between active transport of substances across the cell membrane and passive transport?

What substances are removed from the cell and how?

Elements of the correct answer

1. In prokaryotes, the cell lacks a nucleus, mitochondria, the Golgi apparatus and the endoplasmic reticulum.

2. Prokaryotes do not have true sexual reproduction.

Answer yourself

Why are mature erythrocytes or platelets not classified as prokaryotic cells, despite the absence of nuclei in them?

Why are viruses not classified as independent organisms?

Why are eukaryotic organisms more diverse in structure and complexity?

Elements of the correct answer

1. By the chromosome set of an animal, you can determine its type.

2. By the chromosome set of an animal, you can determine its sex.

3. According to the chromosome set of an animal, it is possible to determine the presence or absence of hereditary diseases.

Answer yourself

Does every cell in a multicellular organism have chromosomes? Prove your answer with examples.

How and when can you see chromosomes in a cell?

Elements of the correct answer

The structural elements of the Golgi complex are:

1) tubules;
2) cavities;
3) bubbles.

Answer yourself

What is the structure of a chloroplast?

What is the structure of a mitochondrion?

What must be contained in mitochondria so that they can synthesize proteins?

Prove that both mitochondria and chloroplasts can multiply.

Elements of the correct answer

Note the differences in:

1) the nature of metabolism;
2) terms of life;
3) reproduction.

Answer yourself

How will the transplantation of a nucleus from another organism affect a single-celled organism?

Elements of the correct answer

1. The presence of a double membrane with characteristic nuclear pores, which ensures the connection of the nucleus with the cytoplasm.

2. The presence of nucleoli, in which RNA is synthesized and ribosomes are formed.

3. The presence of chromosomes, which are the hereditary apparatus of the cell and ensure nuclear division.

Answer yourself

Which cells do not contain nuclei?

Why do non-nuclear prokaryotic cells reproduce, but non-nuclear eukaryotic cells do not?

Elements of the correct answer

1. Most cells are similar in basic structural elements, vital properties and the process of division.

2. Cells differ from each other in the presence of organelles, specialization in the functions performed, and the intensity of metabolism.

Answer yourself

Give examples of the correspondence of the structure of a cell to its function.

Give examples of cells with different levels of metabolic intensity.

Elements of the correct answer

1. As a result of synthesis, more complex substances are formed than those that have reacted; the reaction proceeds with the absorption of energy.

2. During the decay, simpler substances are formed than those that have reacted; The reaction proceeds with the release of energy.

Answer yourself

What are the functions of enzymes in metabolic reactions?

Why are more than 1000 enzymes involved in biochemical reactions?

Elements of the correct answer

1. Light energy is converted into chemical and thermal energy.

2. All transformations occur in the thylakoids of the gran chloroplasts and in their matrix (in plants); in other photosynthetic pigments (in bacteria).

Answer yourself

What happens in the light phase of photosynthesis?

What happens in the dark phase of photosynthesis?

Why is it difficult to experimentally detect the process of plant respiration in the daytime?

Elements of the correct answer

1. The code "triplet" means that each of the amino acids is encoded by three nucleotides.

2. The code is "unambiguous" - each triplet (codon) encodes only one amino acid.

3. The code "degenerate" means that each amino acid can be encoded by more than one codon.

Answer yourself

Why do we need "punctuation marks" between genes and why are they not inside genes?

What does the concept of "universality of the DNA code" mean?

What is the biological meaning of transcription?

Elements of the correct answer

1. Examples of organisms in which alternation of generations occurs can be mosses, ferns, jellyfish and others.

2. In plants, the gametophyte and sporophyte change. Jellyfish have alternating polyp and medusa stages.

Answer yourself

What are the main differences between mitosis and meiosis?

What is the difference between the terms "cell cycle" and "mitosis"?

Elements of the correct answer

1. Isolated body cells living in an artificial environment are called cell culture (or cell culture).

2. Cell cultures are used to obtain antibodies, drugs, as well as to diagnose diseases.

Elements of the correct answer

1. Interphase is necessary for the storage of substances and energy in preparation for mitosis.

2. In the interphase, the hereditary material is doubled, which subsequently ensures its uniform distribution among the daughter cells.

Answer yourself

Are the gametes produced by an organism the same or different in their genetic composition? Bring evidence.

Which organisms have an evolutionary advantage - haploid or diploid? Bring evidence.

Level C2 tasks

Elements of the correct answer

Mistakes were made in sentences 2, 3, 5.

In sentence 2, notice one of the non-macro elements.

In sentence 3, one of the listed elements is erroneously assigned to microelements.

In sentence 5, the element that performs the named function is erroneously indicated.

Elements of the correct answer

Mistakes were made in sentences 1, 4, 6.

In sentence 1, the monomers of the protein molecule are incorrectly indicated.

Sentence 4 incorrectly indicates the chemical bonds that support the secondary structure of the protein.

Sentence 6 incorrectly indicates the chemical bonds that support the tertiary structure of the protein.