Total mass of DNA molecules. Ready-made solutions to cytology problems. Was there evolution?

2.3 Molecular weight, content and localization of DNA and RNA in the cell; types of DNA and RNA

The molecular mass of DNA is determined mainly by hydrodynamic and electron microscopic methods, although this can be done by measuring the light scattering of DNA solutions and some other methods.

The hydrodynamic method is based on linear dependence DNA sedimentation constant, determined by ultracentrifugation of DNA solutions, from its molecular weight, which can be established using a calibration curve or calculated using the formula: 0.445lgM=1.819+lg(s20?w-2.7), where s20?w is the sedimentation constant given by extrapolation to infinite dilution (s), standard temperature (20 C) and water viscosity (w).

The electron microscopic method for determining the molecular weight of DNA is based on measuring the length of elongated DNA molecules. It is known that per 0.1 nm of the length of its molecule there is a mass equal to 197 Da. By multiplying this value by the experimentally found length, the molecular mass value is found. The molecular weight of eukaryotic DNA is higher than that of prokaryotic DNA (for example, in one of the chromosomes of the fruit fly Drosophila it reaches 7.9 x 10 10). In addition, mitochondria and chloroplasts contain circular DNA molecules with a molecular weight of 10 6 -10 7. The DNA of these organelles is called cytoplasmic; it makes up approximately 0.1% of all cellular DNA.

Depending on the location of DNA in the cell, nuclear, mitochondrial, chloroplast, centriole and episomal DNA are distinguished. Nuclear DNA in eukaryotes sharply prevails over the DNA of other subcellular structures. Thus, from 0.5 * 10 6 to 5 * 10 16 g of DNA was found in mitochondria, in chloroplasts - from 10? 16 to 150*10? 16, and in centrioles - 2 * 10 16 g, which is several percent of nuclear DNA. The same ratio is found in the DNA content in the bacterial chromosome and episomes - extrachromosomal, independently replicating determinants of heredity in microorganisms that ensure the transfer of genetic information, for example, about resistance to antibiotics (otherwise they are called R-factors, i.e. resistance factors). The question of the existence of extrachromosomal DNA, transported or communication DNA, cytoplasmic membrane DNA, and finely dispersed supercoiled DNA is discussed. Based on their functional purpose, they are divided into ribosomal DNA (rDNA) and satellite DNA (stDNA).

In addition to intracellular DNA, there is also DNA that is part of viruses and bacteriophages. Its amount in viruses is significantly lower than in bacterial cells (thousandths of a picogram).

The molecular weights of RNA are determined by the same methods as DNA, but, in addition, electrophoresis in a polyacrylamide gel is used, since the range of RNA in the gel is inversely proportional to their molecular weights. As for the content and localization of RNA in cells, it is neither uniform nor stable: in cells where intensive protein biosynthesis occurs, the RNA content is several times higher than that of DNA (for example, in the liver of a rat there is 4 times more RNA than DNA ), but where protein synthesis is low, the ratio of DNA and RNA can be reversed (for example, in the lungs of a rat, RNA is 2 times less than DNA).

By functional significance and molecular weights, as well as according to localization in the cellular contents, RNA is divided into the following types.

1. Transfer RNAs (tRNAs) are distinguished by relatively low molecular weights (25-30 thousand daltons). tRNAs make up 10% of all RNAs. These RNAs are localized in the cell hyaloplasm, nuclear sap, structureless part of chloroplasts and mitochondria and acquire a specific configuration in the form of a clover leaf. They encode amino acids and transfer them to the ribosomal apparatus of the cell during protein biosynthesis.

2. Ribosomal RNA (rRNA) is characterized mainly by large molecular weights (1-1.5 million daltons), the molecules are large, they contain up to 5,000 nucleotides. They are localized in ribosomes, being their structural basis and performing various functions in them (formation of the active center of the ribosome; ensuring the interaction of rRNA and tRNA).

3. Information, or matrix, RNA (mRNA) have molecular weights that vary widely (from 300,000 to 4*10 6). mRNAs are synthesized in the nucleus during the transcription process at a specific section of the DNA molecule (gene). The function of mRNA is to transfer genetic information about the structure of a protein from DNA to the site of protein synthesis, to ribosomes.

4. Viral RNAs are characterized by diverse and high molecular weights, mostly within the range of several million daltons. They are components of viral and phage ribonucleoproteins and carry all the information necessary for the reproduction of the virus in the host cells.

In modern literature, the question of the advisability of separating several more types of RNA into separate categories is discussed: nuclear, chromosomal, mitochondrial, low-molecular-weight regulatory, antisense.

Nitrogen and its compounds

The bulk of nitrogen on Earth is in a gaseous state and makes up over 3/4 of the atmosphere (78.09% by volume, or 75.6% by mass). Almost on our planet the supply of nitrogen is inexhaustible - 3.8*10^15 tons. Nitrogen is a rather inert element...

Alkaloids of plant materials

Squirrels

Proteins are high molecular weight compounds. These are polymers consisting of hundreds and thousands of amino acid residues - monomers. Accordingly, the molecular weight of proteins is in the range of 10,000-1,000,000. So, as part of ribonuclease (enzyme...

Bismuth and its compounds in nature

Bismuth is a sedentary water migrant and its concentration in groundwater is about 20 µg/dm3, in sea ​​waters- 0.02 μg/dm3. At such concentrations, bismuth does not have a negative effect on water quality...

Coking of hard coals

The organic mass of coal is formed by compounds based on carbon, hydrogen, oxygen, sulfur and nitrogen. Carbon Carbon is the main element of fossil coals...

Determination of iron +2 content in ceramic samples

Clay is a colloidal plastic material secondary origin, formed as a result of the decay and decomposition of certain types of primary rocks...

Calculation of a packed batch distillation column for separating a binary mixture methyl alcohol– benzene

The mass of the apparatus is calculated by the formula, (6.38) where is the mass of the body, kg; - mass of the nozzle, kg; - bottom mass, kg; - weight of the cover, kg; - mass of maximum load on supports, kg. Let's calculate the components of the sum. The mass of the body is (6...

Structure and deformation-strength properties of isoprene rubber

The structure of isoprene rubber is considered at two levels: molecular and supramolecular. The molecular level is characterized by the structure of repeating units, the structure of the polymer chain...

Quantity Content, kg/h, kmol/h, mass. mole fractions shares 105 26283 250.3 0.169732 0.034593 3.63 18 123477 6859.8 0.797400 0.948062 17.07 34 4067.8 119.6 0.026269 0.016529 0.56 244 548.2 2 .25 0.003540 0.000311 0.08 139 467.6 3.36 0.003020 0.000464 0.06 16 1.8 0.113 0...

Technological calculation of an absorber for purifying hydrocarbon gas from regenerated hydrogen sulfide aqueous solution diethanolamine

MPa, K, MPa, K 0.75 4.605 190.55 0.0104 3.4538 142.91 0.00780 0.1 4.875 305.43 0.0986 0.4875 30.54 0.00986 0.08 4.248 369.82 0.15 24 0.3398 29.59 0.01219 0.06 3.795 425.16 0.2010 0.2277 25.51 0.01206 0.01 9.000 373.6 0.1000 0.0900 3.74 0.00100 U 1.00 4.6 232.29 0...

Technological calculation of an absorber for purifying hydrocarbon gas from hydrogen sulfide with a regenerated aqueous solution of diethanolamine

Coefficients in the formula for calculating the enthalpy of an ideal gas Enthalpies, kJ/kg A B C D 0.5372 154.15 15.12 0.0519 56.62 650.3 349.3 0.1343 58.65 23.63 0.4139 56.15 445.7 59.9 0.1576 33.65 26.31 0.5380 35.58 390.9 61.6 0.1558 34.72 26.08 0.5455 39.22 393.4 61, 3 0.0152 87...

Technological calculation of an absorber for purifying hydrocarbon gas from hydrogen sulfide with a regenerated aqueous solution of diethanolamine

Coefficients in the formula for calculating the enthalpy of an ideal gas Enthalpies, kJ/kg A B C D 0.5459 154.15 15.12 0.0519 56.62 650.3 355 0.1365 58.65 23.63 0.4139 56, 15 445.7 60.84 0.1604 33.65 26.31 0.5380 35.58 390.9 62.7 0.1581 34.72 26.08 0.5455 39.22 393.4 62.2 0...

Toxic effect of thallium

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Characteristics of the adsorption process

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Chemical bonding and structure of matter

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The DNA molecule consists of two strands forming a double helix. Its structure was first deciphered by Francis Crick and James Watson in 1953.

At first, the DNA molecule, consisting of a pair of nucleotide chains twisted around each other, gave rise to questions about why it had this particular shape. Scientists call this phenomenon complementarity, which means that only certain nucleotides can be found opposite each other in its strands. For example, adenine is always opposite thymine, and guanine is always opposite cytosine. These nucleotides of the DNA molecule are called complementary.

Schematically it is depicted like this:

T - A

C - G

These pairs form a chemical nucleotide bond, which determines the order of amino acids. In the first case it is a little weaker. The connection between C and G is stronger. Non-complementary nucleotides do not form pairs with each other.

About the building

So, the structure of the DNA molecule is special. It has this shape for a reason: the fact is that the number of nucleotides is very large, and a lot of space is needed to accommodate long chains. It is for this reason that the chains are characterized by a spiral twist. This phenomenon is called spiralization, it allows the threads to shorten by about five to six times.

The body uses some molecules of this type very actively, others rarely. The latter, in addition to spiralization, also undergo such “compact packaging” as superspiralization. And then the length of the DNA molecule decreases by 25-30 times.

What is the “packaging” of a molecule?

The process of supercoiling involves histone proteins. They have the structure and appearance of a spool of thread or a rod. Spiralized threads are wound onto them, which immediately become “compactly packaged” and take up little space. When the need arises to use one or another thread, it is unwound from a spool, for example, a histone protein, and the helix unwinds into two parallel chains. When the DNA molecule is in this state, the necessary genetic data can be read from it. However, there is one condition. Obtaining information is possible only if the structure of the DNA molecule has an untwisted form. Chromosomes that are accessible for reading are called euchromatins, and if they are supercoiled, then they are already heterochromatins.

Nucleic acids

Nucleic acids, like proteins, are biopolymers. The main function is the storage, implementation and transmission of hereditary (genetic information). They come in two types: DNA and RNA (deoxyribonucleic and ribonucleic). The monomers in them are nucleotides, each of which contains a phosphoric acid residue, a five-carbon sugar (deoxyribose/ribose) and a nitrogenous base. The DNA code includes 4 types of nucleotides - adenine (A) / guanine (G) / cytosine (C) / thymine (T). They differ in the nitrogenous base they contain.

In a DNA molecule, the number of nucleotides can be huge - from several thousand to tens and hundreds of millions. Such giant molecules can be examined through an electron microscope. In this case, you will be able to see a double chain of polynucleotide strands, which are connected to each other by hydrogen bonds of the nitrogenous bases of the nucleotides.

Research

During the course of research, scientists discovered that the types of DNA molecules differ in different living organisms. It was also found that guanine of one chain can only bind to cytosine, and thymine to adenine. The arrangement of nucleotides in one chain strictly corresponds to the parallel one. Thanks to this complementarity of polynucleotides, the DNA molecule is capable of doubling and self-reproduction. But first, the complementary chains, under the influence of special enzymes that destroy paired nucleotides, diverge, and then in each of them the synthesis of the missing chain begins. This happens due to the available large quantities free nucleotides in each cell. As a result of this, instead of the “mother molecule”, two “daughter” ones are formed, identical in composition and structure, and the DNA code becomes the original one. This process is a precursor to cell division. It ensures the transmission of all hereditary data from mother cells to daughter cells, as well as to all subsequent generations.

How is the gene code read?

Today, not only the mass of a DNA molecule is calculated - it is also possible to find out more complex data that was previously inaccessible to scientists. For example, you can read information about how an organism uses its own cell. Of course, at first this information is in encoded form and has the form of a certain matrix, and therefore it must be transported to a special carrier, which is RNA. Ribonucleic acid is able to penetrate into the cell through the nuclear membrane and read the encoded information inside. Thus, RNA is a carrier of hidden data from the nucleus to the cell, and it differs from DNA in that it contains ribose instead of deoxyribose, and uracil instead of thymine. In addition, RNA is single-stranded.

RNA synthesis

In-depth analysis of DNA has shown that after RNA leaves the nucleus, it enters the cytoplasm, where it can be integrated as a matrix into ribosomes (special enzyme systems). Guided by the information received, they can synthesize the appropriate sequence of protein amino acids. The ribosome learns from the triplet code which type of organic compound needs to be attached to the forming protein chain. Each amino acid has its own specific triplet, which encodes it.

After the formation of the chain is completed, it acquires a specific spatial form and turns into a protein capable of performing its hormonal, construction, enzymatic and other functions. For any organism it is a gene product. It is from it that all kinds of qualities, properties and manifestations of genes are determined.

Genes

Sequencing processes were primarily developed to obtain information about how many genes a DNA molecule has in its structure. And, although research has allowed scientists to make great progress in this matter, it is not yet possible to know their exact number.

Just a few years ago it was assumed that DNA molecules contain approximately 100 thousand genes. A little later, the figure decreased to 80 thousand, and in 1998, geneticists stated that only 50 thousand genes are present in one DNA, which are only 3% of the total DNA length. But the latest conclusions of geneticists were striking. Now they claim that the genome includes 25-40 thousand of these units. It turns out that only 1.5% of chromosomal DNA is responsible for coding proteins.

The research did not stop there. Parallel team of specialists genetic engineering found that the number of genes in one molecule is exactly 32 thousand. As you can see, it is still impossible to get a definitive answer. There are too many contradictions. All researchers rely only on their results.

Was there evolution?

Despite the fact that there is no evidence of the evolution of the molecule (since the structure of the DNA molecule is fragile and small in size), scientists still made one assumption. Based on laboratory data, they voiced the following version: at the initial stage of its appearance, the molecule had the form of a simple self-replicating peptide, which included up to 32 amino acids found in the ancient oceans.

After self-replication, thanks to the forces of natural selection, molecules acquired the ability to protect themselves from external elements. They began to live longer and reproduce in larger quantities. Molecules that found themselves in the lipid bubble had every chance to reproduce themselves. As a result of a series of successive cycles, lipid bubbles acquired the form of cell membranes, and then - the well-known particles. It should be noted that today any section of a DNA molecule is a complex and clearly functioning structure, all the features of which scientists have not yet fully studied.

Modern world

Recently, scientists from Israel have developed a computer that can perform trillions of operations per second. Today it is the fastest car on Earth. The whole secret is that the innovative device is powered by DNA. Professors say that in the near future, such computers will even be able to generate energy.

A year ago, specialists from the Weizmann Institute in Rehovot (Israel) announced the creation of a programmable molecular computing machine consisting of molecules and enzymes. They replaced silicon microchips with them. To date, the team has made further progress. Now just one DNA molecule can provide a computer with the necessary data and the necessary fuel.

Biochemical “nanocomputers” are not a fiction; they already exist in nature and are manifested in every living creature. But often they are not managed by people. A person cannot yet operate on the genome of any plant in order to calculate, say, the number “Pi”.

The idea of ​​using DNA for storing/processing data first came to the minds of scientists in 1994. It was then that a molecule was used to solve a simple mathematical problem. From that moment on, a number of research groups proposed various projects concerning DNA computers. But here all attempts were based only on the energy molecule. You cannot see such a computer with the naked eye; it looks like a transparent solution of water in a test tube. There are no mechanical parts in it, but only trillions of biomolecular devices - and this is just in one drop of liquid!

Human DNA

People became aware of the type of human DNA in 1953, when scientists were first able to demonstrate to the world a double-stranded DNA model. For this Kirk and Watson received Nobel Prize, since this discovery became fundamental in the 20th century.

Over time, of course, they proved that a structured human molecule can look not only like in the proposed version. After conducting a more detailed DNA analysis, they discovered the A-, B- and left-handed form Z-. Form A- is often an exception, since it is formed only if there is a lack of moisture. But this is only possible if laboratory research, this is anomalous for the natural environment; such a process cannot occur in a living cell.

The B- shape is classic and is known as a double right-handed chain, but the Z- shape is not only twisted in the opposite direction to the left, but also has a more zigzag appearance. Scientists have also identified the G-quadruplex form. Its structure has not 2, but 4 threads. According to geneticists, this form occurs in areas where there is an excess amount of guanine.

Artificial DNA

Today there is already artificial DNA, which is an identical copy of the real one; it perfectly follows the structure of the natural double helix. But, unlike the original polynucleotide, the artificial one has only two additional nucleotides.

Since the dubbing was created based on information obtained from various studies of real DNA, it can also be copied, self-replicating and evolving. Experts have been working on the creation of such an artificial molecule for about 20 years. The result is an amazing invention that can use the genetic code in the same way as natural DNA.

To the four existing nitrogenous bases, geneticists added two additional ones, which were created by chemical modification of natural bases. Unlike natural DNA, artificial DNA turned out to be quite short. It contains only 81 base pairs. However, it also reproduces and evolves.

Replication of a molecule obtained artificially takes place thanks to the polymerase chain reaction, but so far this does not happen independently, but through the intervention of scientists. They independently add the necessary enzymes to the said DNA, placing it in a specially prepared liquid medium.

Final result

The process and final outcome of DNA development can be influenced by various factors, such as mutations. This makes it necessary to study samples of matter so that the analysis result is reliable and reliable. An example is a paternity test. But we can’t help but rejoice that incidents such as mutation are rare. Nevertheless, samples of matter are always rechecked in order to obtain more accurate information based on the analysis.

Plant DNA

Thanks to high technology Sequencing (HTS) has also revolutionized the field of genomics - isolating DNA from plants is also possible. Of course, obtaining high-quality molecular weight DNA from plant material poses some difficulties due to the large number of copies of mitochondria and chloroplast DNA, as well as high level polysaccharides and phenolic compounds. To isolate the structure we are considering in this case, a variety of methods are used.

Hydrogen bond in DNA

The hydrogen bond in the DNA molecule is responsible for the electromagnetic attraction created between a positively charged hydrogen atom that is attached to an electronegative atom. This dipole interaction does not meet the criterion of a chemical bond. But it can occur intermolecularly or in different parts of the molecule, i.e. intramolecularly.

A hydrogen atom attaches to the electronegative atom that is the donor of the bond. An electronegative atom can be nitrogen, fluorine, or oxygen. It - through decentralization - attracts the electron cloud from the hydrogen nucleus to itself and makes the hydrogen atom (partially) positively charged. Since the size of H is small compared to other molecules and atoms, the charge is also small.

DNA decoding

Before deciphering a DNA molecule, scientists first take a huge number of cells. For the most accurate and successful work, about a million of them are needed. The results obtained during the study are constantly compared and recorded. Today, genome decoding is no longer a rarity, but an accessible procedure.

Of course, deciphering the genome of a single cell is an impractical exercise. The data obtained during such studies are of no interest to scientists. But it is important to understand that all currently existing decoding methods, despite their complexity, are not effective enough. They will only allow reading 40-70% of the DNA.

However, Harvard professors recently announced a method through which 90% of the genome can be deciphered. The technique is based on adding primer molecules to isolated cells, with the help of which DNA replication begins. But even this method cannot be considered successful; it still needs to be refined before it can be openly used in science.

“Preparation for Biology Olympiads” - Cumulative effect. Subsequence. Vitamin. Stages of preparing students for Olympiads and intellectual competitions. Are common guidelines. We wish you success in the correspondence stage of the tournament. Formation of students' skills. An interconnected collection of populations. Method biological research. Male ostrich incubating eggs. Cross section. Methods of preparation for Olympiads and intellectual tournaments in biology.

"Ticket in Biology" - A number of differences. Parmelia. Lichens. Cell. Organoid. ATP. Lifeless habitats. A unit of structure and vital activity of organisms. The role of proteins in the body. The role of proteins.

“Typical Unified State Examination tasks in biology” - Correspondence between the characteristic of an organism and the kingdom. B5 Comparison of the structural features and functioning of the body. Generalization and application of knowledge. Comparison of structural features and functioning. Comparison of biological objects. Selection. Ability to work. Typical mistakes when completing USE tasks in biology. Complex. Organisms. Establishing the sequence of biological objects.

“GIA in Biology” - Structure and content of the standard in biology. Distribution of marks on a five-point scale. Number of exams. Oral exam in biology. Distribution diagram of participants' marks. Final certification of students. Recommendations. Control measuring materials 2013. Forms of implementation. State final certification. Preparation for the exam. Problems of final certification in biology. Structure biological education.

“C5 in biology” - The total mass of all DNA molecules. Protein biosynthesis. DNA. Determination of length. Calculation of the number of nucleotides. Determine the number of amino acids. Determine the nucleotide sequence. Solving the problems of part C5. Primary structure of insulin. Molecular biology. Tasks. Molecular mass. Cell division. A section of a DNA molecule chain. Number of nucleotides. How many nucleotides A, T, G are contained in a fragment of a DNA molecule.

“Testing in biology” - Bacterial spores. Test work in biology. How many types of tissues exist in the human and animal body? Sexual parts of a flower. Liver. Formation of organic substances. Numerous organelles. Plants by type of nutrition. Generative organ of a plant. Functions. Carp. Existence of cells. Digestive system. Exoskeleton. Terrestrial vertebrates. Chlorophyll. Animals. Euglena green. Scorpios.

The topics “Molecular Biology” and “Genetics” are the most interesting and complex topics in the “General Biology” course. These topics are studied in both 9th and 11th grades, but there is clearly not enough time in the program to develop the ability to solve problems. However, the ability to solve problems in genetics and molecular biology is provided for by the Standard of Biological Education, and such problems are also part of the Unified State Exam KIM.

To solve problems in molecular biology, you must master the following biological concepts: types of nucleic acids, DNA structure, DNA replication, DNA functions, RNA structure and functions, genetic code, properties genetic code,mutation.

Typical problems introduce the basic techniques of reasoning in genetics, and “story” problems more fully reveal and illustrate the features of this science, making it interesting and attractive for students. The selected tasks characterize genetics as an exact science that uses mathematical methods of analysis. Solving problems in biology requires the ability to analyze factual material, think and reason logically, as well as a certain ingenuity in solving especially difficult and confusing problems.

To consolidate theoretical material on methods and techniques for solving problems, tasks are offered for independent solution, as well as questions for self-control.

Examples of problem solving

Necessary clarifications:

• One step is a complete turn of the DNA helix—a 360 o turn
• One step is 10 nucleotide pairs
• Single step length – 3.4 nm
• The distance between two nucleotides is 0.34 nm
• Molecular mass of one nucleotide – 345 g/mol
• Molecular weight of one amino acid – 120 g/mol
• In a DNA molecule: A+G=T+C (Chargaff’s rule: ∑(A) = ∑(T), ∑(G) = ∑(C), ∑(A+G) =∑(T+C)
• Nucleotide complementarity: A=T; G=C
• DNA chains are held together by hydrogen bonds that form between complementary nitrogenous bases: adenine and thymine are connected by 2 hydrogen bonds, and guanine and cytosine by three.
• On average, one protein contains 400 amino acids;
• Calculation of protein molecular weight:

Where M min is the minimum molecular weight of the protein,
a is the atomic or molecular mass of the component,
c – percentage of the component.

Task No. 1. One of the DNA chains has the nucleotide sequence: AGT ACC GAT ACCT CGA TTT ACG... What is the nucleotide sequence of the second DNA chain of the same molecule. For clarity, you can use the magnetic “alphabet” of DNA (technique of the author of the article).
Solution: According to the principle of complementarity, we complete the second chain (A-T, G-C). It looks like this: TCA TGG CTA TGA GCT AAA TGC.

Task No. 2. The sequence of nucleotides at the beginning of the gene that stores information about the protein insulin begins like this: AAA CAC CTG CTT GTA GAC. Write the sequence of amino acids that begins the insulin chain.
Solution: The task is performed using a genetic code table, in which nucleotides in mRNA (in parentheses - in the original DNA) correspond to amino acid residues.

Task No. 3. The larger of the two insulin protein chains (the so-called B chain) begins with the following amino acids: phenylalanine-valine-asparagine- glutamic acid-histidine-leucine. Write the sequence of nucleotides at the beginning of the section of the DNA molecule that stores information about this protein.

because One amino acid can be encoded by several triplets; the exact structure of mRNA and DNA sections cannot be determined; the structure may vary. Using the principle of complementarity and the genetic code table, we get one of the options:

Task No. 4. The gene section has the following structure, consisting of a sequence of nucleotides: CGG CGC TCA AAA TCG... Indicate the structure of the corresponding protein section, information about which is contained in this gene. How will the removal of the fourth nucleotide from the gene affect the structure of the protein?

Solution (for convenience, we use a tabular form for recording the solution):

When the fourth nucleotide, C, is removed from the gene, noticeable changes will occur - the number and composition of amino acids in the protein will decrease:

Task No. 5. The tobacco mosaic virus (RNA-containing virus) synthesizes a protein section with the amino acid sequence: Ala - Tre - Ser - Glu - Met-. Under the influence of nitrous acid (a mutagenic factor), cytosine is converted into uracil as a result of deamination. What structure will the tobacco mosaic virus protein region have if all cytidyl nucleotides undergo the indicated chemical transformation?

Solution (for convenience, we use a tabular form for recording the solution): Using the principle of complementarity and the genetic code table we get:

Task No. 6. With Fancomi syndrome (disorder of education bone tissue) in the patient's urine, amino acids are excreted, which correspond to codons in i-RNA: AUA GUTS AUG UCA UUG GUU AUU. Determine the excretion of which amino acids in the urine is characteristic of Fancomi syndrome, if you have healthy person urine contains the amino acids alanine, serine, glutamic acid, glycine.

Solution (for convenience, we use a tabular form for recording the solution): Using the principle of complementarity and the genetic code table we get:

Thus, in the urine of a sick person, only one amino acid (serine) is the same as in a healthy person, the rest are new, and three characteristic of a healthy person are absent.

Task No. 7. Chain A of bovine insulin contains alanine in the 8th link, and threonine in the horse, and serine and glycine in the 9th link, respectively. What can be said about the origin of insulins?

Solution (for ease of comparison, we use a tabular form of recording the solution): Let's see which triplets in mRNA encode the amino acids mentioned in the problem statement.

Because amino acids are encoded by different triplets, triplets that differ minimally from each other are taken. In this case, in the horse and bull, the amino acids in the 8th and 9th links are changed as a result of the replacement of the first nucleotides in triplets and -RNA: guanine is replaced by adenine (or vice versa). In double-stranded DNA this would be equivalent to replacing pairs C-G to T-A (or vice versa).
Consequently, the differences between the A chains of bovine and horse insulin are due to transitions in the section of the DNA molecule encoding the 8th and 9th links of the A chain of bovine and horse insulins.

Task No. 7. Studies have shown that mRNA contains 34% guanine, 18% uracil, 28% cytosine and 20% adenine. Determine the percentage composition of nitrogenous bases in the DNA section that is the template for this mRNA.
Solution (for convenience, we use a tabular form for recording the solution): The percentage of nitrogenous bases is calculated based on the principle of complementarity:

The total of A+T and G+C in the semantic chain will be: A+T=18%+20%=38%; G+C=28%+34%=62%. In the antisense (non-coding) chain, the total indicators will be the same, only the percentage of individual bases will be the opposite: A+T=20%+18%=38%; G+C=34%+28%=62%. In both chains there will be equal amounts of complementary base pairs, i.e. adenine and thymine - 19% each, guanine and cytosine 31% each.

Task No. 8. On a fragment of one DNA strand, the nucleotides are arranged in the sequence: A–A–G–T–C–T–A–C–G–T–A–T. Determine the percentage of all nucleotides in this DNA fragment and the length of the gene.

Solution:

1) we complete the second thread (according to the principle of complementarity)

2) ∑(A +T+C+G) = 24, of which ∑(A) = 8 = ∑(T)

∑(G) = 4 = ∑(C)

3) the DNA molecule is double-stranded, so the length of the gene is equal to the length of one chain:

12 × 0.34 = 4.08 nm

Task No. 9. Cytidyl nucleotides account for 18% of the DNA molecule. Determine the percentage of other nucleotides in this DNA.

Solution:

1) because C = 18%, then G = 18%;
2) A+T accounts for 100% – (18% +18%) = 64%, i.e. 32% each

Task No. 10. There are 880 guanidyl nucleotides found in a DNA molecule, which account for 22% of the total number of nucleotides in this DNA. Determine: a) how many other nucleotides are there in this DNA? b) what is the length of this fragment?

Solution:

1) ∑(G) = ∑(C)= 880 (this is 22%); The share of other nucleotides accounts for 100% - (22% + 22%) = 56%, i.e. 28% each; To calculate the number of these nucleotides, we make up the proportion:

22% – 880
28% – x, hence x = 1120

2) to determine the length of DNA, you need to find out how many total nucleotides are contained in 1 chain:

(880 + 880 + 1120 + 1120) : 2 = 2000
2000 × 0.34 = 680 (nm)

Task No. 11. A DNA molecule with a relative molecular weight of 69,000 is given, of which 8,625 are adenyl nucleotides. Find the number of all nucleotides in this DNA. Determine the length of this fragment.

Solution:

1) 69,000: 345 = 200 (nucleotides in DNA), 8625: 345 = 25 (adenyl nucleotides in this DNA), ∑(G+C) = 200 – (25+25) = 150, i.e. there are 75 of them;
2) 200 nucleotides in two chains, which means there are 100 in one. 100 × 0.34 = 34 (nm)

Task No. 12. What is heavier: a protein or its gene?

Solution: Let x be the number of amino acids in a protein, then the mass of this protein is 120x, the number of nucleotides in the gene encoding this protein is 3x, the mass of this gene is 345 × 3x. 120x< 345 × 3х, значит ген тяжелее белка.

Task No. 13. Human blood hemoglobin contains 0.34% iron. Calculate the minimum molecular weight of hemoglobin.

Solution: M min = 56: 0.34% 100% = 16471

Task No. 14. Human serum albumin has a molecular weight of 68400. Determine the number of amino acid residues in the molecule of this protein.

Solution: 68400: 120 = 570 (amino acids in an albumin molecule)

Task No. 15. The protein contains 0.5% glycine. What is the minimum molecular weight of this protein if M glycine = 75.1? How many amino acid residues are there in this protein?

Solution: M min = 75.1: 0.5% 100% = 15020; 15020: 120 = 125 (amino acids in this protein)

Tasks for independent work

1. The DNA molecule split into two chains. one of them has the structure: TAG ACC GGT ACA CGT GGT GAT TCA... What structure will the second DNA molecule have when the indicated chain is completed to a complete double-stranded molecule?
2. The polypeptide chain of one animal protein has the following beginning: lysine-glutamine-threonine-alanine-alanine-alanine-lysine-... What nucleotide sequence does the gene corresponding to this protein begin with?
3. A section of the protein molecule has the following sequence of amino acids: glutamine-phenylalanine-leucine-tyrosine-arginine. Determine one of the possible nucleotide sequences in a DNA molecule.
4. A section of a protein molecule has the following sequence of amino acids: glycine-tyrosine-arginine-alanine-cysteine. Determine one of the possible nucleotide sequences in a DNA molecule.
5. One of the chains of ribonuclease (pancreatic enzyme) consists of 16 amino acids: Glu-Gly-asp-Pro-Tyr-Val-Pro-Val-Pro-Val-Gis-phen-Phen-Asn-Ala-Ser-Val. Determine the structure of the DNA region encoding this part of the ribonuclease.
6. The DNA gene fragment has the following nucleotide sequence GTC CTA ACC GGA TTT. Determine the sequence of mRNA nucleotides and amino acids in the polypeptide chain of a protein.
7. The DNA gene fragment has the following nucleotide sequence TCG GTC AAC TTA GCT. Determine the sequence of mRNA nucleotides and amino acids in the polypeptide chain of a protein.
8. The DNA gene fragment has the following nucleotide sequence TGG ACA GGT TTC GTA. Determine the sequence of mRNA nucleotides and amino acids in the polypeptide chain of a protein.
9. Determine the order of amino acids in a section of a protein molecule if it is known that it is encoded by the following sequence of DNA nucleotides: TGA TGC GTT TAT GCG CCCC. How will the protein change if the 9th and 13th nucleotides are chemically removed?
10. The coding strand of DNA has the nucleotide sequence: TAG TsGT TTC TCG GTA. How will the structure of a protein molecule change if the sixth nucleotide in the DNA chain is doubled. Explain the results.
11. The coding strand of DNA has the nucleotide sequence: TAG TTC TCG AGA. How will the structure of a protein molecule change if the eighth nucleotide in the DNA chain is doubled. Explain the results.
12. Under influence mutagenic factors in the gene fragment: TsAT TAG GTA TsGT TCG, the second triplet was replaced by the ATA triplet. Explain how the structure of the protein molecule will change.
13. Under the influence of mutagenic factors in the gene fragment: AGA TAG GTA CGT TCG, the fourth triplet was replaced by the ACC triplet. Explain how the structure of the protein molecule will change.
14. A fragment of an mRNA molecule has the following nucleotide sequence: GCA UGU AGC AAG CGC. Determine the sequence of amino acids in a protein molecule and its molecular weight.
15. A fragment of an mRNA molecule has the following nucleotide sequence: GAG CCA AAU ACU UUA. Determine the sequence of amino acids in a protein molecule and its molecular weight.
16. A DNA gene consists of 450 base pairs. What is the length, molecular weight of the gene and how many amino acids are encoded in it?
17. How many nucleotides does a DNA gene contain if it encodes 135 amino acids? What is the molecular weight of this gene and its length?
18. A fragment of one DNA strand has the following structure: GGT ACG ATG TCA AGA. Determine the primary structure of the protein encoded in this chain, amount (%) various types nucleotides in two chains of the fragment and its length.
19. What is the molecular mass of the gene and its length if it encodes a protein with a molecular mass of 1500 g/mol?
20. What is the molecular mass of the gene and its length if it encodes a protein with a molecular mass of 42000 g/mol?
21. The protein molecule contains 125 amino acids. Determine the number of nucleotides in the mRNA and DNA gene, as well as the number of tRNA molecules that took part in the synthesis of this protein.
22. The protein molecule contains 204 amino acids. Determine the number of nucleotides in the mRNA and DNA gene, as well as the number of tRNA molecules that took part in the synthesis of this protein.
23. 145 tRNA molecules took part in the synthesis of the protein molecule. Determine the number of nucleotides in the mRNA, DNA gene and the number of amino acids in the synthesized protein molecule.
24. 128 tRNA molecules took part in the synthesis of the protein molecule. Determine the number of nucleotides in the mRNA, DNA gene and the number of amino acids in the synthesized protein molecule.
25. The fragment of the mRNA chain has the following sequence: YYY UGG UAU CCC AAC UGU. Determine the nucleotide sequence on DNA, t-RNA anticodons, and the amino acid sequence corresponding to the DNA gene fragment.
26. The fragment of the mRNA chain has the following sequence: GUU GAA CCG UAU GCU. Determine the nucleotide sequence on DNA, t-RNA anticodons, and the amino acid sequence corresponding to the DNA gene fragment.
27. The mRNA molecule contains 13% adenyl, 27% guanyl and 39% uracil nucleotides. Determine the ratio of all types of nucleotides in the DNA from which this mRNA was transcribed.
28. The mRNA molecule contains 21% cytidyl, 17% guanyl and 40% uracil nucleotides. Determine the ratio of all types of nucleotides in the DNA from which this mRNA was transcribed
29. The mRNA molecule contains 21% guanyl nucleotides; how many cytidyl nucleotides are contained in the coding strand of the DNA section?
30. If the chain of the DNA molecule from which genetic information is transcribed contained 11% adenyl nucleotides, how many uracil nucleotides will be contained in the corresponding segment of i-RNA?

Used Books.

1. Bolgova I.V. Collection of problems on general biology with solutions for applicants to universities - M.: Onyx Publishing House LLC: "Publishing House." World and Education, 2008.
2. Vorobiev O.V. Biology lessons using information technology. 10th grade. Toolkit with electronic application – M.: Planeta, 2012.
3. Cherednichenko I.P. Biology. Interactive didactic materials. 6-11 grades. Methodological manual with electronic interactive application. – M.: Planeta, 2012.
4. Internet links:

a protein consisting of 400 amino acids is encoded. average mass of nucleotides in a DNA molecule

3 In one DNA molecule, timini accounts for 18%; determine the % ratio of other nucleotides in the DNA molecule

this molecule contains 51 amino acids, and the linear length of one nucleotide in nucleic acid is 3.4 angstroms?

2. What is the mass of the part of the DNA molecule that encodes the insulin molecule, if it is known that this molecule contains 51 amino acids, and the average molecular weight of one nucleotide is 345 a. O. m.

2%. Determine the relative molecular mass of a fragment of a molecule, taking into account that the relative molecular mass of one nucleotide is 354, and the number of all types of nucleotides in a given DNA molecule.

2. What signs of variability are transmitted to the offspring (modification, mutation)?
3. What changes when mutations occur (genotype, phenotype)?
4. Are genotype or phenotype traits inherited?
5. What variability is characterized by the following characteristics: occur suddenly, can be dominant or recessive, beneficial or harmful, inherited, repeated (mutational, modification)?
6. Where do mutations occur (in chromosomes, in DNA molecules, in one pair of nucleotides, in several nucleotides)?
7. In what case does the mutation manifest itself phenotypically (in any, in a homozygous organism, in a heterozygous organism)?
8. What is the role of mutations in the evolutionary process (increasing variability, adaptation to the environment, self-improvement of the organism)?
9. What does the phenotype depend on (the genotype, the environment, nothing else)?
10. What determines the range of variability in an organism’s characteristics ( environment, genotype)?
11. Signs of what variability are expressed in the form of a variation series and a variation curve (mutation, modification)?
12. Which signs have a narrow reaction rate (qualitative, quantitative), which are more flexible (qualitative, quantitative)?
13. Which form of natural selection in a population leads to the formation of new species (driving, stabilizing), which - to the preservation of species characteristics (driving, stabilizing)?