Cell theory, its main provisions, role in the formation of the modern natural science picture of the world. Development of knowledge about the cell. Cellular structure organisms, the similarity of the structure of the cells of all organisms is the basis of the unity of the organic world, evidence of the kinship of living nature.
A cell is a unit of structure, vital activity, growth and development of organisms. Diversity of cells. Comparative characteristics cells of plants, animals, bacteria, fungi.
The structure of pro- and eukaryotic cells. The relationship between the structure and functions of the parts and organelles of a cell is the basis of its integrity. Metabolism: energy and plastic metabolism, their relationship. Enzymes, their chemical nature, role in metabolism. Stages energy metabolism. Fermentation and respiration. Photosynthesis, its significance, cosmic role. Phases of photosynthesis. Light and dark reactions of photosynthesis, their relationship. Chemosynthesis.
Biosynthesis of protein and nucleic acids. Matrix nature of biosynthesis 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. Life cycle cells: interphase and mitosis. Mitosis is the division of somatic cells. Meiosis. Phases of mitosis and meiosis. 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.
Author of the article - D. A. Solovkov, Candidate of Biological Sciences
Types of Cytology Problems
Cytology tasks that appear in the Unified State Exam can be divided into seven main types. The first type involves determining the percentage of nucleotides in DNA and is most often found in Part A of the exam. The second includes calculation 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.
Cytology tasks of types 3, 4 and 5 are devoted to working with the genetic code table, and also require the applicant to have knowledge of the processes of transcription and translation. Such tasks make up the majority of C5 questions in the Unified State Exam.
Problems of types 6 and 7 appeared in the Unified State Exam relatively recently, and they can also be encountered by an applicant in Part C. The sixth type is based on knowledge about changes in the genetic makeup of a cell during mitosis and meiosis, and the seventh type tests the student’s mastery of material on dissimilation in a eukaryotic cell .
Below are solutions to problems of all types and examples for independent work. The appendix provides 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 adenine in one chain there is always thymine in the other and vice versa (A-T and T-A); opposite cytosine is 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).
Task: the 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 quantities are equal, then C=G=.
Solving problems of the second type
Basic information:
- Amino acids necessary for protein synthesis are delivered to ribosomes using 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: t-RNA molecules were involved 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 encodes 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 occurs 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: AAGGCTACTGTTG. Build mRNA 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-UGC-AAU. Using the genetic code table, 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 a tRNA that is complementary to the nucleotides in an iRNA 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: the mRNA fragment has the following structure: GAUGAGUATSUUCAA. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized.
Solution: we break the mRNA into triplets GAU-GAG-UAC-UUC-AAA and determine the sequence of amino acids using the genetic code table: asp-glu-tyr-phen-lys. This fragment contains triplets, so t-RNA will participate in the synthesis. Their anticodons are determined by the complementarity rule: TsUA, TsUC, AUG, AAG, UUU. Also, according to the rule of complementarity, we determine the DNA fragment (by mRNA!!!): CTATCTCATGAAGTTT.
Solving problems of the fifth type
Basic information:
- The tRNA molecule is synthesized on DNA according to the rule of complementarity.
- Don't forget that RNA contains uracil instead of thymine.
- An anticodon is a sequence of three nucleotides 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. Determine 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 task, use the genetic code table.
Solution: we determine the composition of the t-RNA molecule: ААУЦГГЦУАГГЦ and find the third triplet - this is CUA. This anticodon is complementary to the triplet mRNA - GAC. It encodes the amino acid asp, which is carried by this tRNA.
Solving problems of the sixth type
Basic information:
- The two main ways of cell division are mitosis and meiosis.
- Changes in the genetic makeup of a cell during mitosis and meiosis.
Problem: 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 division of meiosis.
Solution: According to the 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 are there in dissimilation, where do they take place, what are they? chemical reactions pass during each stage.
Task: glucose molecules have entered into dissimilation. Determine the amount of ATP after glycolysis, after the energy stage and the total effect of dissimilation.
Solution: write the equation of glycolysis: = 2PVK + 4H + 2ATP. Since one molecule of glucose produces PVK and 2ATP molecules, 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
- T=, G=C= by .
- amino acids, triplets, nucleotides.
- triplet, amino acids, tRNA molecules.
- mRNA: CCG-AGA-UCG-AAG. Amino acid sequence: pro-arg-ser-lys.
- DNA fragment: CGATTACAAGAAATG. T-RNA anticodons: CGA, UUA, CAA, GAA, AUG. Amino acid sequence: ala-asn-val-ley-tyr.
- tRNA: UCG-GCU-GAA-CHG. Anticodon GAA, codon i-RNA - CUU, transferred amino acid - leu.
- . Genetic set:
- Since one molecule of glucose produces PVK and 2ATP molecules, 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.
- PVK molecules entered the Krebs cycle, therefore, glucose molecules disintegrated. The amount of ATP after glycolysis - molecules, after the energy stage - molecules, the total effect of dissimilation of ATP molecules.
So, this article presents the main types of problems in cytology that an applicant may encounter in the Unified State Exam in biology. We hope that the problem options and their solutions will be useful to everyone when preparing for the exam. Good luck!
Mishnina Lidiya Alexandrovna
biology teacher
MBOU secondary school No. 3 Akbulak village
Class 11
Preparation for the Unified State Exam: solving problems in cytology
In the methodological recommendations for improving the teaching of biology, developed on the basis of an analysis of the difficulties of graduates at the Unified State Exam in 2014, the authors G.S. Kalinova, R.A. Petrosova, noted low level performing tasks to determine the number of chromosomes and DNA in different phases of mitosis or meiosis.
The tasks are actually not so difficult as to cause serious difficulties. What should be taken into account when preparing graduates on this issue?
Solving cytological problems requires knowledge not only of mitosis and meiosis, their phases and the events occurring in them, but also the mandatory possession of information about the structure and functions of chromosomes, the amount of genetic material in the cell.
Therefore, we begin our preparation by reviewing material about chromosomes. We focus on the fact that chromosomes are nucleoprotein structures in the nucleus of a eukaryotic cell.
They contain about 99% of the total DNA of the cell; the rest of the DNA is located in other cellular organelles, determining cytoplasmic inheritance. DNA in the chromosomes of eukaryotes is in complex with the main proteins - histones and with non-histone proteins, which provide complex packaging of DNA in chromosomes and regulation of its ability to synthesize ribonucleic acids (RNA) - transcription.
The appearance of chromosomes changes significantly different stages cell cycle and how compact formations with a characteristic chromosome morphology are clearly distinguishable in a light microscope only during the period of cell division.
At the metaphase stage of mitosis and meiosis, chromosomes consist of two longitudinal copies, which are called sister chromatids and which are formed during DNA replication during the S period of interphase. In metaphase chromosomes, sister chromatids are joined at a primary constriction called the centromere. The centromere is responsible for the separation of sister chromatids into daughter cells during division
The complete set of chromosomes in a cell, characteristic of a given organism, is called a karyotype. In any cell of the body of most animals and plants, each chromosome is represented twice: one of them is received from the father, the other from the mother during the fusion of the nuclei of germ cells during the process of fertilization. Such chromosomes are called homologous, and a set of homologous chromosomes is called diploid.
Now you can repeat the material about cell division.
Of the interphase events, we consider only the synthetic period, so as not to scatter the attention of schoolchildren, but to focus only on the behavior of chromosomes.
Let us remember: during the synthetic (S) period, the doubling of genetic material occurs through DNA replication. It occurs in a semi-conservative manner, when the double helix of the DNA molecule diverges into two chains and a complementary chain is synthesized on each of them.
The result is two identical DNA double helices, each consisting of one new and one old DNA strand. The amount of hereditary material doubles, but the number of chromosomes remains the same - the chromosome becomes bichromatid (2n4c).
Consider the behavior of chromosomes during mitosis:
- In prophase, metaphase - 2p 4s - since cell division does not occur;
- During anaphase, chromatids separate, the number of chromosomes doubles (chromatids become independent chromosomes, but for now they are all in one cell) 4n 4c;
- in telophase 2n2c (single-chromatid chromosomes remain in the cells).
Repeat meiosis:
- In prophase 1, metaphase 1, anaphase 1 - 2p 4s - since cell division does not occur;
- in telophase - p2c remains, since after the divergence of homologous chromosomes, a haploid set remains in the cells, but the chromosomes are bichromatid;
- In prophase 2, metaphase 2, as well as telophase 1 - p2s;
- Special attention turn to anaphase 2, since after chromatid divergence the number of chromosomes increases by 2 times (chromatids become independent chromosomes, but for now they are all in one cell) 2n 2c;
- in telophase 2 - ps (single-chromatid chromosomes remain in the cells.
Only now, when children are theoretically prepared, can we move on to solving problems.
A typical mistake in preparing graduates: we try to solve problems right away without repeating the material. What happens: children decide with a teacher, but the solution occurs at the level of mechanical memorization, without comprehension. Therefore, when they are given a similar task in an exam, they fail to cope with it. I repeat: there was no comprehension in solving problems.
Let's move on to practice.
We use a selection of problems from the website “I will solve the Unified State Exam” by Dmitry Gushchin. What makes this resource attractive is that there are practically no errors, and the answer standards are well written.
Let's look at problem C 6 No. 12018.
The chromosome set of wheat somatic cells is 28.
Determine the chromosome set and the number of DNA molecules in one of the ovule cells before the onset of meiosis, in anaphase of meiosis 1 and in anaphase of meiosis 2. Explain what processes occur during these periods and how they affect the change in the number of DNA and chromosomes.
Response elements:
The cells of the ovule contain a diploid set of chromosomes - 28 (2n2c).
Before the start of meiosis - (2n4c) 28 xp, 56 DNA
In anaphase of meiosis 1: (2n4c = n2c+n2c) - 28 xp, 56 DNA.
Meiosis 2 involves 2 daughter cells with a haploid set of chromosomes (n2c) - 14 chromosomes, 28 DNA.
In anaphase of meiosis 2: (2n2с= nc+nc) - 28 chromosomes, 28 DNA
The task is complex; how can one help a graduate comprehend its solution?
One option: we draw the phases of meiosis and show all the manipulations with chromosomes.
Action algorithm:
- Read the problem carefully, define the task, write down the phases in which you need to indicate the amount of genetic material
a) Before the start of meiosis
b) In anaphase of meiosis 1
c) In anaphase of meiosis 2
- Make drawings for each designated phase of meiosis and explain what you did.
Let me clarify: we don’t use drawings, we make them ourselves. This operation works for comprehension ( although we lose in aesthetics, we win in the end!)
1. Before the start of meiosis
Let me explain: meiosis is preceded by interphase, in interphase DNA doubling occurs, therefore the number of chromosomes is 2n, the number of DNA is 4c.
2. In anaphase of meiosis 1
Let me explain: in anaphase of meiosis 1, chromosomes diverge to the poles, i.e. From each pair of homologous chromosomes, only one gets into the daughter cell. The chromosome set becomes haploid, but each chromosome consists of two chromatids. Since cell division has not yet occurred and all chromosomes are in one cell, the chromosome formula can be written as: 2n4c (n2c+n2c) 28 xp, 56 DNA (14 xp 28 DNA + 14 xp 28DNA)
3) In anaphase of meiosis 2
Anaphase of meiosis 2 occurs after the first (reduction) division. Set of chromosomes in a p2c cell. During anaphase of meiosis, the 2 centromeres connecting sister chromatids divide and the chromatids, as in mitosis, become independent chromosomes. The number of chromosomes increases and becomes equal to 2n2c. And again, since cell division has not yet occurred and all chromosomes are in one cell, the chromosome set can be written as follows: 2n2c (nc + nc) 28 xp, 28 DNA (14 xp 14 DNA + 14 xp 14 DNA).
- Write down the answer. (we have it listed above)
To sum it up: Solving problems of this type does not require a pursuit of quantity; here it is important to achieve an understanding of the logic of the solution and knowledge of the behavior of chromosomes at each phase of division.
Resources used:
- FIPI " Guidelines on some aspects of improving the teaching of biology" ed. G.S. Kalinova, R.A. Petrosova. Moscow, 2014
- Biology. General patterns 10th grade: textbook for educational institutions/ V.B.Zakharov, S.G.Mamontov, N.I.Sonin - Moscow: Bustard Publishing House, 2011.
- I will solve the Unified State Exam. http://bio.reshuege.ru/
In the lesson we will learn the history of the origin of cytology, remember the concept of a cell, and consider the contribution that various scientists made to the development of cytology.
All living beings, with the exception of vi-ru-s, are made up of cells. But for scientists of the past, the cellular structure of living organisms was not as obvious as for you and me. Science that studies the cell, cytology, was formed only in the middle of the 19th century. Without knowledge of where life comes from, that it appears in the smallest unit, right up to the Middle Ages theories have emerged that, for example, frogs come from dirt, and mice are born in dirty underwear (Fig. 2).
Rice. 2. Theories of the Middle Ages ()
“The dirty laundry of middle-century science” was first “sewn” in 1665. English nature-is-py-ta-tel Ro- Bert Hooke (Fig. 3).
Rice. 3. Robert Hooke ()
For the first time he looked at and described the shells of plant cells. And already in 1674, his Dutch coll-le-ha An-to-ni van Leeuwen-hoek (Fig. 4) was the first to be seen under a self-made mik -ro-sko-po-some of the simplest and individual cells of animals, such as erythro-cy-ts and sperm-to-zos -yes.
Rice. 4. Anthony van Leeuwenhoek ()
The research of Le-ven-gu-ka has become so fan-ta-sti-che-ski-mi that in 1676 year, the London Co-ro-lion-society, where he sent the results of his research, very much in them for-with-me. The existence of one-cell organs and blood cells, for example, does not fit into the framework of that wherever science is.
It took several centuries to comprehend the results of the work of the Dutch scientist. Only by the middle of the 19th century. German scientist Theodor Schwann, based on the work of his colleague Ma-tti-a-sa Schlei-de-na (Fig. 5 ), forming the basic principles of cellular theory, which we still use to this day.
Rice. 5. Theodor Schwann and Matthias Schleiden ()
Schwann realized that the cells of plants and animals have a common principle of structure, because they form the same in what way; all cells are self-sustaining, and any organism is a collection of life-individuals groups of cells (Fig. 6).
Rice. 6. Red blood cells, cell division, DNA molecule ()
Further research into scientific positions will form the basic principles of modern times cellular theory:
- A cell is a universal structural unit of life.
- Cells multiply by dividing (cell from cell).
- Cells are stored, re-ra-ba-you-va-yut, re-a-li-zu-yut and re-y-y-yat to the successive information-form-ma -tion.
- A cell is the most powerful bio-system, from a certain structural level of the or- ga-ni-za-tion of living ma-te-rii.
- Multi-cell-precise orga-niz-we are a complex of inter-acting systems of different cells that provide chi-va-yu-shchih or-ga-low-mu growth, development, metabolism and energy.
- The cells of all organisms are similar to each other in structure, composition and function.
The cells are different. They can differ in structure, shape and function (Fig. 7).
Rice. 7. Cell diversity ()
Among them there are free-living cells, which behave like individuals of populations and species, like their own organisms. ga-bottom-we. Their vitality depends not only on how the internal cellular structures, or-ga, work -but-and-y. They themselves need to get their own food, move around in the environment, multiply, that is, act act like small, but quite self-sufficient individuals. There are a lot of such free-loving ones. They are included in all kingdoms of cellular living nature and inhabit all environments of life on our planet. In a multi-cell-precise organ-ga-bottom, the cell is part of it, from the cells tissues and organ-ga are formed -us.
The size of the cells can be very different - from one tenth of a micron to 15 centimeters - this is the size of a country's egg , representing one cell, and the weight of this cell is half a kilogram. And this is not the limit: the eggs of di-no-saurs, for example, could reach a length of as much as 45 centimeters (Fig. 8) .
Rice. 8. Dinosaur egg ()
Usually, in multi-cell organiza- tions, different cells perform different functions. Cells similar in structure, located nearby, united by intercellular substance and purpose -needed to perform certain functions in the organization, form tissues (Fig. 9).
Rice. 9. Tissue formation ()
Life is a lot of or-ga-niz-ma dependent on how weakly the cells work, entering members in its composition. This is why the cells do not compete with each other; on the contrary, there is co-operation and specialization of their functions Possible or-ga-niz-mu to survive in those si-tu-a-tsi-yah, in which single cells do not you-li-v-va- ut. In complex multi-cell or- ganisms - plants, animals and humans - the cells of the or-ga-ni- zones in the fabric, fabrics - in the organs, organs - in the system of the organs. And each of these systems works to ensure the existence of the whole organization.
Despite all the different shapes and sizes, cells of different types are similar to each other. Processes such as respiration, bio-synthesis, metabolism occur in cells regardless of whether they are -but-kle-toch-ny-mi or-ga-niz-ma-mi or are included in the composition of many-kle-exactly-no-beings. Each cell eats food, draws energy from it, and receives energy from waste products. of a society that maintains the stability of its own hi-mi-che-s-co-sta-va and reproduces itself, that is, she carries out all the processes on which her life depends.
All this allows us to consider the cell as a special unit of living matter, as an elemental living system ( Fig. 10).
Rice. 10. Schematic drawing of a cell ()
All living creatures, from in-fu-zo-ria to an elephant or a whale, the largest mammal today wow, they're made of cells. The only difference is that in-fu-zo-rii are the most-storable bio-systems, consisting of one cell, and the cells of the whale are or-ga-ni-zo-va-ny and inter-and-mo-connected as parts of a large 190-ton whole. The condition of the entire or-ga-niz-ma depends on how its parts, that is, cells, function.
Bibliography
- Mamontov S.G., Zakharov V.B., Agafonova I.B., Sonin N.I. Biology. General patterns. - Bustard, 2009.
- Ponomareva I.N., Kornilova O.A., Chernova N.M. Fundamentals of general biology. 9th grade: Textbook for 9th grade students of general education institutions / Ed. prof. I.N. Ponomareva. - 2nd ed., revised. - M.: Ventana-Graf, 2005
- Pasechnik V.V., Kamensky A.A., Kriksunov E.A. Biology. Introduction to general biology and ecology: Textbook for grade 9, 3rd ed., stereotype. - M.: Bustard, 2002.
- Krugosvet.ru ().
- Uznaem-kak.ru ().
- Mewo.ru ().
Homework
- What does cytology study?
- What are the main provisions of cell theory?
- How are cells different?
D. A. Solovkov, Candidate of Biological Sciences
This collection of tasks contains all the main types of cytology tasks found in the Unified State Exam and is intended primarily for self-study applicant to solve task C5 on the exam. For convenience, the tasks are grouped according to the main sections and topics included in the biology program (section “Cytology”). Self-test answers are provided at the end.
Examples of problems of the first type
Examples of problems of the second type
Examples of problems of the third type
- A fragment of one of the DNA chains has the following structure: AAGTCGTGCTCAG. Build mRNA on it and determine the sequence of amino acids in a fragment of a protein molecule (use the genetic code table for this).
- A fragment of one of the DNA chains has the following structure: CCATATTCCGGAT. Build mRNA on it and determine the sequence of amino acids in a fragment of a protein molecule (use the genetic code table for this).
- A fragment of one of the DNA chains has the following structure: AGTTTTCTTGGGCAA. Build mRNA on it and determine the sequence of amino acids in a fragment of a protein molecule (use the genetic code table for this).
- A fragment of one of the DNA chains has the following structure: GATTACCTAGTT. Build mRNA on it and determine the sequence of amino acids in a fragment of a protein molecule (use the genetic code table for this).
- A fragment of one of the DNA chains has the following structure: CTATCCTGCTGTC. Build mRNA on it and determine the sequence of amino acids in a fragment of a protein molecule (use the genetic code table for this).
- A fragment of one of the DNA chains has the following structure: AAGCTACAGACC. Build mRNA on it and determine the sequence of amino acids in a fragment of a protein molecule (use the genetic code table for this).
- A fragment of one of the DNA chains has the following structure: GGTGCCGGAAAG. Build mRNA on it and determine the sequence of amino acids in a fragment of a protein molecule (use the genetic code table for this).
- A fragment of one of the DNA chains has the following structure: CCTGTAATTTCG. Build mRNA on it and determine the sequence of amino acids in a fragment of a protein molecule (use the genetic code table for this).
Examples of problems of the fourth type
- The mRNA fragment has the following structure: GAUGAGUATSUUCAA. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
- The mRNA fragment has the following structure: TsGAGGUAUUCTTSUGG. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
- The mRNA fragment has the following structure: UGUUCAAAAAGGAAGG. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
- The mRNA fragment has the following structure: CCGCCAACACGCGAGC. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
- The mRNA fragment has the following structure: ACAGUGGGCCAACCCC. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
- The mRNA fragment has the following structure: GATSAGATSUCAAGUTSU. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
- The mRNA fragment has the following structure: UGCATSUGAACGTCGUA. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
- The mRNA fragment has the following structure: GCAGGCCAGUAUAUAU. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
- The mRNA fragment has the following structure: GCUAAUGUUCUUUAC. Determine the tRNA anticodons and the amino acid sequence encoded in this fragment. Also write the fragment of the DNA molecule on which this mRNA was synthesized (use the genetic code table for this).
Examples of problems of the fifth type
- The DNA fragment has the following nucleotide sequence TATGGGGCTATTG. Determine 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 task, use the genetic code table.
- The DNA fragment has the following nucleotide sequence CAAGATTTTTGTT. Determine 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 task, use the genetic code table.
- The DNA fragment has the following nucleotide sequence GCCAATCCCTGA. Determine 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 task, use the genetic code table.
- The DNA fragment has the following nucleotide sequence TGTCCATCAAAAC. Determine 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 task, use the genetic code table.
- The DNA fragment has the following nucleotide sequence CATGAAAATGAT. Determine 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 task, use the genetic code table.
Examples of problems of the sixth type
Examples of problems of the seventh type
Appendix I Genetic code (mRNA)
| First base | Second base | Third base | |||
| U | C | A | G | ||
| U | Hairdryer | Ser | Shooting gallery | Cis | U |
| Hairdryer | Ser | Shooting gallery | Cis | C | |
| Lei | Ser | - | - | A | |
| Lei | Ser | - | Three | G | |
| C | Lei | About | Gies | Arg | U |
| Lei | About | Gies | Arg | C | |
| Lei | About | Gln | Arg | A | |
| Lei | About | Gln | Arg | G | |
| A | Ile | Tre | Asn | Ser | U |
| Ile | Tre | Asn | Ser | C | |
| Ile | Tre | Liz | Arg | A | |
| Meth | Tre | Liz | Arg | G | |
| G | Shaft | Ala | Asp | Gli | U |
| Shaft | Ala | Asp | Gli | C | |
| Shaft | Ala | Glu | Gli | A | |
| Shaft | Ala | Glu | Gli | G |
Answers
- A=. G=C=.
- A=. G=C=.
- C=. A=T=.
- C=. A=T=.
- G=. A=T=.
- G=. A=T=.
- amino acids, triplets, nucleotides.
- amino acids, triplets, nucleotides.
- triplet, amino acid, tRNA molecule.
- triplet, amino acids, tRNA molecules.
- triplets, amino acids, tRNA molecules.
- i-RNA: UUC-GCA-CGA-GUC. Amino acid sequence: fen-ala-arg-val.
- i-RNA: GGU-AUA-GGC-CUA. Amino acid sequence: gly-ile-gly-ley.
- mRNA: UCA-AAG-CCG-GUU. Amino acid sequence: ser-lys-pro-val.
- mRNA: CUA-AUG-GAU-CAA. Amino acid sequence: leu-met-asp-gln.
- mRNA: GAU-AGG-CGA-CAG. Amino acid sequence: asp-arg-arg-gln.
- i-RNA: UUC-GAU-GUC-UGG. Amino acid sequence: phen-asp-val-tri.
- i-RNA: CCA-CGG-CCU-UUC. Amino acid sequence: pro-arg-pro-phen.
- mRNA: GGG-CAU-UUA-AGC. Amino acid sequence: gly-his-leu-ser.
- DNA fragment: CTATTSATGAAGTTT. T-RNA anticodons: TsUA, TsUC, AUG, AAG, UUU. Amino acid sequence: asp-glu-tyr-phen-lys.
- DNA fragment: GCTCCATAAGGGACC. tRNA anticodons: GCU, CCA, UAA, GGG, ACC. Amino acid sequence: arg-gly-ile-pro-tri.
- DNA fragment: ACAAGTTTATCTTTC. tRNA anticodons: ACA, AGU, UAU, CCU, UCC. Amino acid sequence: cis-ser-ile-gly-arg.
- DNA fragment: GGCGTTTGTGCGCTTG. TRNA anticodons: GGC, GUU, GUG, CGC, UCG. Amino acid sequence: pro-gln-gis-ala-ser.
- DNA fragment: TGTCACGGTTGGGA. tRNA anticodons: UGU, TsAC, CGG, UUG, GGA. Amino acid sequence: tre-val-ala-asn-pro.
- DNA fragment: TsTGTCTGGAGTTCAGA. T-RNA anticodons: TsUG, UtsU, GAG, Uts, Aga. Amino acid sequence: asp-arg-leu-lys-ser.
- DNA fragment: ACTGTGACTTTGCGCAT. tRNA anticodons: ACG, UGA, TsUU, GCH, TsAU. Amino acid sequence: cis-tre-glu-arg-val.
- DNA fragment: CGTCCGGGTCAAATA. tRNA anticodons: CGU, CCG, GUC, AAU, AUA. Amino acid sequence: ala-gly-gln-leu-tyr.
- DNA fragment: CGATTACAAGAAATG. T-RNA anticodons: CGA, UUA, CAA, GAA, AUG. Amino acid sequence: ala-asn-val-ley-tyr.
- tRNA: AUA-CCC-GAU-AAC. Anticodon GAU, codon i-RNA - CUA, transferred amino acid - leu.
- tRNA: GUU-CUA-AAA-CAA. Anticodon AAA, mRNA codon - UUU, transferred amino acid - phen.
- tRNA: CGG-UUU-AGG-ATSU. Anticodon AGG, i-RNA codon - UCC, transferred amino acid - ser.
- tRNA: ACA-GGU-AGU-UUG. Anticodon AGU, mRNA codon - UCA, transferred amino acid - ser.
- tRNA: GUA-CUU-UUA-CUA. Anticodon UUA, mRNA codon - AAU, transferred amino acid - asn.
- . Genetic set:
- . Genetic set:
- . Genetic set:
- . Genetic set:
- . Genetic set:
- . Genetic set:
- . Genetic set:
- . Genetic set:
- Since one molecule of glucose produces PVK and ATP molecules, 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.
- Since one molecule of glucose produces PVK and ATP molecules, 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.
- Since one molecule of glucose produces PVK and ATP molecules, 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.
- PVA molecules entered the Krebs cycle, therefore, glucose molecules disintegrated. The amount of ATP after glycolysis - molecules, after the energy stage - molecules, the total effect of dissimilation of ATP molecules.
- PVA molecules entered the Krebs cycle, therefore, glucose molecules disintegrated. The amount of ATP after glycolysis - molecules, after the energy stage - molecules, the total effect of dissimilation of ATP molecules.
- PVA molecules entered the Krebs cycle, therefore, glucose molecules disintegrated. The amount of ATP after glycolysis - molecules, after the energy stage - molecules, the total effect of dissimilation of ATP molecules.
- PVA molecules entered the Krebs cycle, therefore, glucose molecules disintegrated. The amount of ATP after glycolysis - molecules, after the energy stage - molecules, the total effect of dissimilation of ATP molecules.
