Molecular genetics – branch of genetics that deals with the study of heredity at the molecular level.

Nucleic acids. DNA replication. Matrix synthesis reactions

Nucleic acids (DNA, RNA) were discovered in 1868 by the Swiss biochemist I.F. Misher. Nucleic acids are linear biopolymers consisting of monomers - nucleotides.

DNA - structure and functions

The chemical structure of DNA was deciphered in 1953 by the American biochemist J. Watson and the English physicist F. Crick.

General structure of DNA. The DNA molecule consists of 2 chains that are twisted into a spiral (Fig. 11), one around the other and around a common axis. DNA molecules can contain from 200 to 2x10 8 base pairs. Along the helix of the DNA molecule, adjacent nucleotides are located at a distance of 0.34 nm from each other. A complete turn of the helix includes 10 base pairs. Its length is 3.4 nm.

Rice. 11 . DNA structure diagram (double helix)

Polymerism of the DNA molecule. A DNA molecule - a bioploimer - consists of complex compounds - nucleotides.

The structure of a DNA nucleotide. The DNA nucleotide consists of 3 links: one of the nitrogenous bases (adenine, guanine, cytosine, thymine); deoxysyribose (monosaccharide); remainder phosphoric acid(Fig. 12).

There are 2 groups of nitrogenous bases:

purine - adenine (A), guanine (G), containing two benzene rings;

pyrimidine - thymine (T), cytosine (C), containing one benzene ring.

DNA is made up of the following types of nucleotides: adenine (A); guanine (G); cytosine (C); thymine (T). The names of nucleotides correspond to the names of the nitrogenous bases that make up their composition: adenine nucleotide nitrogenous base adenine; guanine nucleotide nitrogenous base guanine; cytosine nucleotide nitrogenous base cytosine; thymine nucleotide nitrogenous base thymine.

Joining two strands of DNA into one molecule

Nucleotides A, G, C and T of one chain are connected, respectively, with nucleotides T, C, G and A of another chain hydrogen bonds. Two hydrogen bonds are formed between A and T, and three hydrogen bonds are formed between G and C (A=T, G≡C).

Base pairs (nucleotides) A ​​- T and G - C are called complementary, that is, mutually corresponding. complementarity is the chemical and morphological correspondence of nucleotides each other in paired strands of DNA.

5 ’ 3 ’

1 2 3

3’ 5’

Rice. 12 Section of the DNA double helix. The structure of the nucleotide (1 - phosphoric acid residue; 2 - deoxyribose; 3 - nitrogenous base). Connection of nucleotides using hydrogen bonds.

Chains in the DNA molecule antiparallel, i.e., directed in opposite directions, so that the 3' end of one strand is opposite the 5' end of the other strand. The genetic information in DNA is written from the 5' end to the 3' end. This strand is called sense DNA,

because that's where the genes are. The second thread - 3'–5' serves as a standard for storing genetic information.

The ratio between the number of different bases in DNA was established by E. Chargaff in 1949. Chargaff found that in DNA of various species the amount of adenine is equal to the amount of thymine, and the amount of guanine is equal to the amount of cytosine.

E. Chargaff's rule:

in a DNA molecule, the number of A (adenine) nucleotides is always equal to the number of T (thymine) nucleotides or the ratio ∑ A to ∑ T=1. The sum of G (guanine) nucleotides is equal to the sum of C (cytosine) nucleotides or the ratio of ∑ G to ∑ C=1;

the sum of purine bases (A + G) is equal to the sum of pyrimidine bases (T + C) or the ratio of ∑ (A + G) to ∑ (T + C) \u003d 1;

Method of DNA synthesis - replication. Replication is the process of self-doubling of the DNA molecule, carried out in the nucleus under the control of enzymes. Self-doubling of the DNA molecule occurs based on complementarity- strict correspondence of nucleotides to each other in paired DNA chains. At the beginning of the replication process, the DNA molecule unwinds (despiralizes) in a certain area (Fig. 13), while hydrogen bonds are released. On each of the chains formed after the rupture of hydrogen bonds, with the participation of the enzyme DNA polymerase, a daughter strand of DNA is synthesized. The material for synthesis is free nucleotides contained in the cytoplasm of cells. These nucleotides line up complementary to the nucleotides of the two parent DNA strands. DNA polymerase enzyme attaches complementary nucleotides to the DNA template strand. For example, for a nucleotide A template chain polymerase adds a nucleotide T and, accordingly, to the G nucleotide, the C nucleotide (Fig. 14). Crosslinking of complementary nucleotides occurs with the help of an enzyme DNA ligases. Thus, two daughter strands of DNA are synthesized by self-duplication.

The resulting two DNA molecules from one DNA molecule are semi-conservative model, since they consist of the old parent and new daughter chains and are an exact copy of the parent molecule (Fig. 14). The biological meaning of replication lies in the exact transfer of hereditary information from the parent molecule to the child.

Rice. 13 . Despiralization of a DNA molecule by an enzyme

1

Rice. 14 . Replication - the formation of two DNA molecules from one DNA molecule: 1 - a daughter DNA molecule; 2 - maternal (parental) DNA molecule.

The DNA polymerase enzyme can only move along the DNA strand in the 3' –> 5' direction. Since the complementary strands in the DNA molecule are directed in opposite directions, and the DNA polymerase enzyme can move along the DNA strand only in the 3'->5' direction, the synthesis of new strands proceeds antiparallel ( according to the principle of anti-parallelism).

Location of DNA. DNA is contained in the cell nucleus, in the matrix of mitochondria and chloroplasts.

The amount of DNA in the cell is constant and is 6.6x10 -12 g.

DNA functions:

Storage and transmission in a number of generations of genetic information to molecules and - RNA;

Structural. DNA is the structural basis of chromosomes (the chromosome is 40% DNA).

DNA species specificity. The nucleotide composition of DNA serves as a species criterion.

RNA, structure and functions.

General structure.

RNA is a linear biopolymer consisting of a single polynucleotide chain. Distinguish between primary and secondary structures of RNA. The primary structure of RNA is a single-stranded molecule, while the secondary structure is cross-shaped and is characteristic of t-RNA.

Polymerism of the RNA molecule. An RNA molecule can be from 70 nucleotides to 30,000 nucleotides. The nucleotides that make up RNA are as follows: adenyl (A), guanyl (G), cytidyl (C), uracil (U). In RNA, a thymine nucleotide is replaced by a uracil nucleotide (U).

The structure of the RNA nucleotide.

The RNA nucleotide includes 3 links:

nitrogenous base (adenine, guanine, cytosine, uracil);

monosaccharide - ribose (in ribose there is oxygen at each carbon atom);

phosphoric acid residue.

Method of RNA synthesis - transcription. Transcription, like replication, is a template synthesis reaction. The matrix is ​​the DNA molecule. The reaction proceeds according to the principle of complementarity on one of the DNA strands (Fig. 15). The transcription process begins with the despiralization of a DNA molecule at a specific site. The transcribed DNA strand has promoter - a group of DNA nucleotides from which the synthesis of an RNA molecule begins. An enzyme binds to a promoter RNA polymerase. The enzyme activates the transcription process. According to the principle of complementarity, the nucleotides coming from the cytoplasm of the cell to the transcribed DNA chain are completed. RNA polymerase activates the alignment of nucleotides in one chain and the formation of an RNA molecule.

There are four stages in the process of transcription: 1) binding of RNA polymerase to a promoter; 2) the beginning of synthesis (initiation); 3) elongation - the growth of the RNA chain, i.e., there is a sequential attachment of nucleotides to each other; 4) termination - completion of mRNA synthesis.

Rice. 15 . Transcription scheme

1 - DNA molecule (double strand); 2 – RNA molecule; 3-codons; 4- promoter.

In 1972, American scientists - virologist H.M. Temin and molecular biologist D. Baltimore discovered reverse transcription on viruses in tumor cells. reverse transcription rewriting of genetic information from RNA to DNA. The process is carried out with the help of an enzyme. reverse transcriptase.

RNA types by function

Messenger or messenger RNA (i-RNA or mRNA) transfers genetic information from the DNA molecule to the site of protein synthesis - the ribosome. Synthesized in the nucleus with the participation of the enzyme RNA polymerase. It makes up 5% of all types of cell RNA. mRNA includes from 300 nucleotides to 30,000 nucleotides (the longest chain among RNA).

Transfer RNA (t-RNA) transports amino acids to the site of protein synthesis, the ribosome. It has the shape of a cross (Fig. 16) and consists of 70 - 85 nucleotides. Its amount in the cell is 10-15% of the cell's RNA.

Rice. sixteen. Scheme of the structure of t-RNA: A-D - pairs of nucleotides connected by hydrogen bonds; E - the place of attachment of the amino acid (acceptor site); E - anticodon.

3. Ribosomal RNA (r-RNA) is synthesized in the nucleolus and is part of the ribosomes. Includes approximately 3000 nucleotides. It makes up 85% of the cell's RNA. This type of RNA is found in the nucleus, in ribosomes, on the endoplasmic reticulum, in chromosomes, in the mitochondrial matrix, and also in plastids.

Fundamentals of Cytology. Solution of typical tasks

Task 1

How many thymine and adenine nucleotides are contained in DNA if 50 cytosine nucleotides are found in it, which is 10% of all nucleotides.

Solution. According to the rule of complementarity in the DNA double strand, cytosine is always complementary to guanine. 50 cytosine nucleotides make up 10%, therefore, according to the Chargaff rule, 50 guanine nucleotides also make up 10%, or (if ∑C = 10%, then ∑G = 10%).

The sum of a pair of nucleotides C + G is 20%

The sum of a pair of nucleotides T + A \u003d 100% - 20% (C + G) \u003d 80%

In order to find out how many thymine and adenine nucleotides are in DNA, you need to make the following proportion:

50 cytosine nucleotides → 10%

X (T + A) → 80%

X \u003d 50x80: 10 \u003d 400 pieces

According to the Chargaff rule, ∑A= ∑T, therefore ∑A=200 and ∑T=200.

Answer: the number of thymine, as well as adenine nucleotides in DNA, is 200.

Task 2

Thymine nucleotides in DNA make up 18% of the total number of nucleotides. Determine the percentage of other types of nucleotides contained in DNA.

Solution.∑T=18%. According to the Chargaff rule, ∑T=∑A, therefore, adenine nucleotides also account for 18% (∑A=18%).

The sum of the T + A base pair is 36% (18% + 18% = 36%). For a pair of nucleotides Gi C accounts for: G + C \u003d 100% -36% \u003d 64%. Since guanine is always complementary to cytosine, their content in DNA will be equal,

i.e. ∑ G= ∑C=32%.

Answer: the content of guanine, like cytosine, is 32%.

Task 3

20 cytosine DNA nucleotides make up 10% of the total number of nucleotides. How many adenine nucleotides are in a DNA molecule?

Solution. In a double strand of DNA, the amount of cytosine is equal to the amount of guanine, therefore, their sum is: C+G=40 nucleotides. Find the total number of nucleotides:

20 cytosine nucleotides → 10%

X (total number of nucleotides) → 100%

X=20x100:10=200 pieces

A + T \u003d 200 - 40 \u003d 160 pieces

Since adenine is complementary to thymine, their content will be equal,

i.e. 160 pieces: 2=80 pieces, or ∑A=∑T=80.

Answer: There are 80 adenine nucleotides in a DNA molecule.

Task 4

Add the nucleotides of the right DNA strand if the nucleotides of its left strand are known: AGA - TAT - GTG - TCT

Solution. The construction of the right DNA chain according to a given left chain is carried out according to the principle of complementarity - strict correspondence of nucleotides to each other: adenone - thymine (A-T), guanine - cytosine (G-C). Therefore, the nucleotides of the right DNA chain should be as follows: TCT - ATA - CAC - AGA.

Answer: nucleotides of the right DNA strand: TCT - ATA - CAC - AGA.

Task 5

Write down the transcription if the transcribed DNA chain has the following nucleotide order: AGA - TAT - THT - TCT.

Solution. The i-RNA molecule is synthesized according to the principle of complementarity on one of the strands of the DNA molecule. We know the order of the nucleotides in the transcribed DNA strand. Therefore, it is necessary to build a complementary strand of mRNA. It should be remembered that instead of thymine, the RNA molecule includes uracil. Hence:

DNA chain: AGA - TAT - TGT - TCT

i-RNA chain: UCU - AUA - ACA - AGA.

Answer: i-RNA nucleotide sequence is as follows: UCU - AUA - ACA -AGA.

Task 6

Write down the reverse transcription, i.e. build a fragment of a double-stranded DNA molecule according to the proposed mRNA fragment, if the mRNA chain has the following nucleotide sequence:

GCG – ACA – UUU – UCG – CSU – ASU – AGA

Solution. Reverse transcription is the synthesis of a DNA molecule based on the genetic code of mRNA. The i-RNA encoding the DNA molecule has the following order of nucleotides: GCG - ACA - UUU - UCG - CGU - AGU - AGA. The DNA chain complementary to it: CHC - TGT - AAA - AGC - HCA - TCA - TCT. Second strand of DNA: GCH-ACA-TTT-TCG-CGT-AGT-AGA.

Answer: as a result of reverse transcription, two chains of the DNA molecule were synthesized: CHC - THT - AAA - AGC - HCA - TCA and GCH - ACA - TTT - TCH - CHT - AGT - AGA.

Genetic code. protein biosynthesis.

Gene- a section of a DNA molecule containing genetic information about the primary structure of one specific protein.

Exon-intron structure of a geneeukaryote

promoter- stretch of DNA (up to 100 nucleotides long) to which the enzyme attaches RNA polymerase required for transcription;

2) regulatory area– zone affecting gene activity;

3) structural part of a gene- genetic information about the primary structure of the protein.

A sequence of DNA nucleotides that carries genetic information about the primary structure of a protein - exon. They are also part of the mRNA. A sequence of DNA nucleotides that does not carry genetic information about the primary structure of a protein – intron. They are not part of the mRNA. In the course of transcription, with the help of special enzymes, copies of introns are cut out from mRNA and copies of exons are fused during the formation of an mRNA molecule (Fig. 20). This process is called splicing.

Rice. 20 . Splicing scheme (formation of mature mRNA in eukaryotes)

genetic code - a system of nucleotide sequences in a DNA molecule, or mRNA, that corresponds to the sequence of amino acids in a polypeptide chain.

Properties of the genetic code:

Tripletity(ACA – GTG – GCG…)

The genetic code is triplet, since each of the 20 amino acids is encoded by a sequence of three nucleotides ( triplet, codon).

There are 64 types of nucleotide triplets (4 3 = 64).

Unambiguity (specificity)

The genetic code is unambiguous because each individual triplet of nucleotides (codon) encodes only one amino acid, or one codon always corresponds to one amino acid (table 3).

Multiplicity (redundancy, or degeneracy)

The same amino acid can be encoded by several triplets (from 2 to 6), since there are 20 protein-forming amino acids, and 64 triplets.

Continuity

The reading of genetic information occurs in one direction, from left to right. If one nucleotide falls out, then when reading it, the nearest nucleotide from the neighboring triplet will take its place, which will lead to a change in genetic information.

Versatility

The genetic code is characteristic of all living organisms, and the same triplets code for the same amino acid in all living organisms.

Has start and end triplets(starting triplet - AUG, terminal triplets UAA, UGA, UAG). These types of triplets do not code for amino acids.

Non-overlapping (discreteness)

The genetic code is non-overlapping, since the same nucleotide cannot simultaneously be part of two adjacent triplets. Nucleotides can belong to only one triplet, and if they are rearranged to another triplet, then the genetic information will change.

Table 3 - Table of the genetic code

		Codon bases

Note: Abbreviated names of amino acids are given in accordance with international terminology.

Protein biosynthesis

Protein biosynthesis - type of plastic exchange substances in the cell, occurring in living organisms under the action of enzymes. Protein biosynthesis is preceded by matrix synthesis reactions (replication - DNA synthesis; transcription - RNA synthesis; translation - assembly of protein molecules on ribosomes). In the process of protein biosynthesis, 2 stages are distinguished:

transcription

broadcast

During transcription, the genetic information contained in the DNA located in the chromosomes of the nucleus is transferred to the RNA molecule. Upon completion of the transcription process, mRNA enters the cytoplasm of the cell through pores in the nuclear membrane, is located between 2 subunits of the ribosome, and participates in protein biosynthesis.

Translation is the process of translating the genetic code into a sequence of amino acids. Translation is carried out in the cytoplasm of the cell on ribosomes, which are located on the surface of the EPS (endoplasmic reticulum). Ribosomes are spherical granules with an average diameter of 20 nm, consisting of large and small subunits. The mRNA molecule is located between two subunits of the ribosome. Amino acids, ATP, i-RNA, t-RNA, the enzyme amino-acyl t-RNA synthetase participate in the translation process.

codon- a section of a DNA molecule, or i-RNA, consisting of three consecutive nucleotides, encoding one amino acid.

Anticodon- a section of the t-RNA molecule, consisting of three consecutive nucleotides and complementary to the codon of the m-RNA molecule. Codons are complementary to the corresponding anticodons and are connected to them via hydrogen bonds (Fig. 21).

Protein synthesis begins with start codon AUG. From him the ribosome

moves along the RNA molecule, triplet by triplet. Amino acids come from the genetic code. Their integration into the polypeptide chain on the ribosome occurs with the help of t-RNA. The primary structure of t-RNA (chain) passes into the secondary structure, resembling a cross in shape, and at the same time, the complementarity of nucleotides is preserved in it. In the lower part of t-RNA there is an acceptor site to which an amino acid is attached (Fig. 16). Activation of amino acids is carried out with the help of an enzyme aminoacyl tRNA synthetase. The essence of this process is that this enzyme interacts with amino acids and with ATP. In this case, a triple complex is formed, represented by this enzyme, amino acid and ATP. The amino acid is enriched with energy, activated, acquires the ability to form peptide bonds with the neighboring amino acid. Without the process of amino acid activation, a polypeptide chain cannot be formed from amino acids.

The opposite, upper part of the tRNA molecule contains a triplet of nucleotides anticodon, with the help of which t-RNA is attached to its complementary codon (Fig. 22).

The first t-RNA molecule, with an activated amino acid attached to it, attaches its anticodon to the mRNA codon, and one amino acid appears in the ribosome. Then the second t-RNA is attached with its anticodon to the corresponding codon of the mRNA. At the same time, 2 amino acids are already in the ribosome, between which a peptide bond is formed. The first tRNA leaves the ribosome as soon as it donates an amino acid to the polypeptide chain on the ribosome. Then the 3rd amino acid is attached to the dipeptide, it is brought by the third t-RNA, etc. Protein synthesis stops at one of the terminal codons - UAA, UAG, UGA (Fig. 23).

1 – mRNA codon; codonsUCG-UCG; CUA-CUA; CGU-CGU;

2 – t-RNA anticodon; anticodon GAT - GAT

Rice. 21 . Translation phase: the mRNA codon is attracted to the tRNA anticodon by the corresponding complementary nucleotides (bases)

1. One of the DNA chains has a nucleotide sequence: AGT AND GAT ACT CGA TTT ACH... What is the nucleotide sequence of the second chain of the same molecule?

Solution.According to the principle of complementarity, we complete the second chain (A-T, G-C). It will look like this: TCA TGG CTA TGA GCT AAA THC...

2. The larger of the two insulin protein chains (called the 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.

Solution.Since several triplets can code for one amino acid, it is impossible to determine the exact structure of the mRNA and the DNA region, and the structure can vary. Using the principle of nucleotide complementarity and the table of the genetic code, we get one of the options:

3. The gene section has the following structure, consisting of a nucleotide sequence: CHG CHC TCA AAA TCG... . Specify the structure of the corresponding region of the protein, 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.Using the principle of complementarity of connecting bases by hydrogen bonds and the table of the genetic code, we do everything as in the previous task:

DNA chain	CHG	CGC	TCA		TCG
IRNA	SC	GCH	ATU	uuu	AGC
Protein chain amino acids	Ala - Ala - Ser - Fen - Ser

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

DNA chain	CHG	Hzt	CAA	AAT	CG
IRNA	SC	CGA	SUM	UUA	HZ
Protein chain amino acids	Ala - Apr - Val - Lei -

4. In Fanconi's syndrome (impaired bone formation), amino acids are excreted in the patient's urine, which correspond to codons in mRNA: AUA, HUC, AUG, UCA, UUG, GUU, AUU. Determine the excretion of which amino acids in the urine is characteristic of Fanconi's syndrome, if a healthy person's urine contains the amino acids alanine, serine, glutamic acid and glycine.

Solution.Using the table of the genetic code, we determine the amino acids that are encoded by the indicated triplets. These are isoleucine, valine, methionine, series, leucine, tyrosine, valine, isoleucine. Thus, in the urine of a patient, only one amino acid (series) is the same as in a healthy person, the remaining six are new, and three, characteristic of a healthy person, are absent.

5. Studies have shown that mRNA contains 84% ​​guanine, 18% uracil, 28% cytosine, 20% adenine. Determine the percentage of nitrogenous bases in the DNA region that is the template for this mRNA.

Solution.Obviously, 34% guanine in mRNA in the sense (readable) DNA strand will be 34% cytosine, respectively, 18% uracil - 18% adenine, 28% cytosine - 28% guanine, 20% adenine - 20% thymine (according to the principle of complementarity nucleotide bases). In total, 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-encoded) strand (DNA is a double-stranded molecule), the total indicators will be the same, only the percentage of individual bases will be reversed: A + T \u003d 20% + 18% - 38%, G + C - 34 % + 28% = 62%. In both chains, the pairs of complementary bases will be equally divided, that is, adenine and thymine - 19% each, guanine and cytosine 31% each.

6. Chain A of bovine insulin contains alanine in the 8th link, and threonine in horses, and the 9th link; series and glycine, respectively. What can be said about the origin of insulins?

Solution.Let's see what triplets in mRNA encode the amino acids mentioned in the condition of the problem, and for the convenience of comparison, we will compile a small table:

	Bull	Horse
8th unit of the protein	ALL	TRE
IRNA	HCU	ACU
9th protein link	SER	GLI
IRNA	ASU	GSU

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

Lecture number 2. DNA replication

According to the hypothesis of J. Watson and F. Crick, each of the chains of the DNA double helix serves as a template for the replication of complementary daughter chains. In this case, two daughter double-stranded DNA molecules are formed, identical to the parent, and each of these molecules contains one unchanged strand of the parent DNA. This mechanism of DNA replication, called semi-conservative, was confirmed in experiments on E. coli cells in 1957 by M. Meselson and F. Stahl. The conservative replication method, in which one daughter DNA must contain both original strands, and the second one must consist of two newly synthesized strands, and the dispersive replication mechanism, in which each daughter DNA strand consists of sections of the parent and newly formed DNA, are excluded (Fig. 1, slide 1) .

DIV_ADBLOCK489">

3. the process is symmetrical: both strands of parental DNA serve as templates; it can also be called semi-conservative;

4. elongation of the DNA chain (or its separate fragment) always occurs in the direction from the 5'-end to the 3'-end. This means that the next new nucleotide is added to the 3' end of the growing chain. In addition, since in any DNA molecule the complementary strands are antiparallel, the growing strand is also antiparallel to the template strand. Therefore, the latter is read in the direction 3'→5' (slide 2 and 3).

5. unpaired DNA strand, which serves as a template, and a seed strand, to which new nucleotides are attached;

The replication process is carried out by a complex enzyme complex. During DNA replication in eukaryotes, not one, but a large number of such complexes work at once on each chromosome. That. There are many origins of DNA replication on the chromosome. And DNA duplication does not occur sequentially from one end to the other, but simultaneously in many places at once, which significantly reduces the duration of the process (slide 5). Replication propagates in both directions from each origin of replication, thus forming replication forks. Between the forks, a gradually expanding "bloat" or "eye" appears - these are already replicated sections of DNA. Neighboring bumps eventually merge and the DNA is duplicated.

The enzyme complex works in such a way that one of the two chains it synthesizes grows with some lead compared to the other chain. Accordingly, the first chain is called leading, and the second - lagging. The leading chain is formed by the enzyme complex in the form of a continuous very long fragment. Its length (for example, for spermatogonia) is 1,600,000 nucleotides; the lagging chain is formed in the form of a series of short fragments - approximately 1,500 nucleotides each. This is the so-called. fragments of the Okazaki.

The formation of each DNA fragment is preceded by the synthesis of a short sequence (of 10-15 nucleotides) of RNA primer. The fact is that DNA polymerase (the main enzyme of DNA synthesis) cannot start the process "from scratch", i.e. in the absence of an oligonucleotide sequence. But the RNA synthesis enzyme (RNA polymerase) has such an ability; and this enzyme starts the formation of each new DNA fragment.

Enzymes and proteins involved in DNA synthesis: DNA polymerases, topoisomerases (gyrases), helicases and ligases, primase, ssb proteins. The whole complex, consisting of more than 20 replicative enzymes and factors, is called the DNA replicase system, or replisome.

DNA-dependent DNA polymerases are the key enzymes of the replication process that use the principle of complementarity to build up polynucleotide chains. Prokaryotes have three DNA polymerases: Pol I, Pol II, and Pol III. Pol I and Pol III are involved in DNA replication. DNA polymerase I has polymerase and (3'→5', 5'→3')-exonuclease activity, participates in primer removal, building up the gap formed at the site of the primer, correction of replication errors, and also in DNA repair. There are about 400 molecules of this enzyme in E. coli cells. Pol III carries out reparative DNA synthesis.

The main enzyme catalyzing the biosynthesis of newly formed DNA in prokaryotes is DNA polymerase III (Pol III). It has polymerase and 3'→5'-exonuclease activity; synthesizes the leading and lagging strands of DNA, has a corrective function. The cell contains 10-20 Pol III molecules, it has an increased affinity for the matrix and provides high copying efficiency.

Activation" href="/text/category/aktivatciya/" rel="bookmark">activation of DNA polymerase.

The question arises, why does DNA polymerase III need 2 types of activity: polymerase and 3¢→5¢ exonuclease? The fact is that the accuracy of copying during DNA replication is very high - there is approximately one error per billion base pairs. However, rare tautomeric forms of all four bases appear for a short time in normal DNA. These forms form irregular pairs. For example, the tautomeric form of cytosine pairs with adenine instead of guanine, resulting in a mutation (slide). This means that the high accuracy of replication is determined by the mechanism that performs the correction, i.e., the elimination of such errors. This is where the 3¢→5¢-exonuclease activity of DNA polymerase III comes into play. Having come into contact with a DNA molecule that has an unpaired cytosine with adenine, DNA polymerase III cleaves off (by hydrolysis) any unpaired nucleotides.

There is evidence that DNA polymerase III catalyzes the coupled synthesis of the leading (leading) and lagging DNA strands during replication. DNA polymerases need a primer because they can only attach deoxyribonucleotides to the 3'-OH group. There is 1 primer on the leading strand, and more than one on the lagging strand. DNA polymerase on the lagging strand synthesizes a short fragment in 4 s, and then switches to the synthesis of another (next) fragment in the site of the template strand located at some distance from the first one (slide).

For each short fragment, DNA polymerase requires a primer (primer) with a paired 3¢ end. Primers are synthesized by the enzyme DNA primase, which forms short RNA primers (primers) from ribonucleoside triphosphates, consisting of about 10 nucleotides in eukaryotes (slide). The primers are synthesized at certain intervals on the template of the lagging strand, then they are built up by DNA polymerase, each time starting a new Okazaki fragment. The DNA polymerase molecule continues to grow until it reaches the seed (primer). In order to ensure the continuity of the DNA chain of many such fragments, the DNA repair system comes into play, which removes the RNA primer and replaces it with DNA. The ligase completes the process, connecting the 3¢-end of the new fragment with the 5¢-end of the previous fragment.

The DNA double strand must unwind as the replication fork progresses so that incoming deoxyribonucleoside triphosphates can pair with the parent template strand. However, in normal conditions the DNA double helix is ​​stable; the paired bases are so strongly bonded that temperatures approaching the boiling point of water (90°C) are required to separate two strands of DNA in a test tube. In order for the double helix to open, two types of proteins are needed: helicases and SSB proteins.

Proteins that prepare parental DNA for replication

a) The origins of replication on the DNA molecule have a specific base sequence rich in A-T pairs.

The process begins with the fact that each such sequence binds several molecules of special recognition proteins. In the case of bacteria, such proteins are called DnaA (as the first proteins initiating replication). (Therefore, in the figure, the recognition protein is labeled A.)

One can imagine various reasons why it becomes possible for recognition proteins to interact with origins of replication. Among these reasons:

- the very appearance in the nucleus of recognizing proteins or their specific modification;

- release of replication start points from certain blocking elements;

- the appearance in the nucleus of some third factors necessary for the interaction in question; etc.

The available data support the first option. But in any case, it is clear that here is one of the key links that control the start of replication.

Recognizing proteins, having ensured the binding of the DNA-replicating complex, apparently do not move further along with it along the DNA.

b) One of the "pioneers" is the helicase enzyme (in the figure it is indicated by the letter D). It provides unwinding in the region of the replication fork of the double helix of the parent DNA: the latter is disconnected into single-stranded sections.

This consumes the energy of ATP hydrolysis - 2 ATP molecules per division of 1 pair of nucleotides.

Apparently, at the same time, the displacement of this DNA segment from its connection with histones and other chromosomal proteins also occurs.

c) However, the unwinding of the helix in a certain area creates supercoiling in front of this area.

The fact is that each DNA molecule is fixed in a number of places on the nuclear matrix. Therefore, it cannot rotate freely when unweaving some of its section. This causes supercoiling, and with it, the formation of structural stress that blocks further unwinding of the double helix.

The problem is solved with the help of topoisomerase enzymes (And in Fig.). Evidently, they function on the as yet ununtwisted section of DNA, i.e., where supercoiling occurs. Topoisomerases involved in the unwinding of the double helix in the replication fork. These enzymes change the degree of supercoiling and lead to the formation of a "hinge" that creates conditions for the continuous movement of the replication fork. Two types of topoisomerases have been identified: Type I topoisomerases cut one of the two strands of DNA, causing the end portion of the double helix to turn around the intact strand, and then rejoin the ends of the cut strand. Type II topoisomerases introduce temporary breaks in both complementary strands, change the degree of supercoiling, and then join the broken ends. Topoisomerases help the helicase unwind DNA for its replication. There is also topoisomerase II (bacterial topoisomerase II is called gyrase). This enzyme breaks both DNA chains at once, again transferring the corresponding ends to itself. This makes it even more effective to solve the problem of supercoils when unwinding DNA.

Topoisomerase I breaks one of the DNA strands, transferring its proximal end to itself (Fig.). This allows the distal portion of DNA (from the point of unwinding to the point of break) to rotate around the corresponding link of the whole strand, which prevents the formation of supercoils. Subsequently, the ends of the broken chain close again: one of them is transferred from the enzyme to the other end. So the process of chain breaking by topoisomerase is easily reversible.

Helicases(from lat. helix- helix, protein dnaB), carry out the formation and movement along the DNA helix of the replication fork - a section of the molecule with unraveled chains. These chain-unwinding enzymes use the energy released by the hydrolysis of ATP. Helicases act in conjunction with ssb-proteins that bind to single-stranded regions of the molecule and thereby stabilize the untwisted duplex.

d) So, “supported” by topoisomerases, the helicase enzyme locally unwinds the DNA double helix into two separate strands.

Special SSB proteins immediately bind to each of these threads. The latter have an increased affinity for single-stranded DNA regions and stabilize them in this state.

Note that in this way these proteins differ from histones, which bind primarily to double-stranded DNA regions.

polymerization enzymes

a) A special protein acts as a primase activator (AP in the figure). After that, primase (P), using the corresponding section of single-stranded DNA as a template, synthesizes a short RNA primer, or primer.

b) Next, DNA polymerases come into play. There are 5 different DNA polymerases known in eukaryotes. Of these, β- and ε-polymerases are involved in DNA repair, γ-polymerase in mitochondrial DNA replication, and α- and δ-polymerases in nuclear DNA replication.

At the same time, according to some assumptions, α-polymerase is associated with both primase and δ-polymerase, and the latter, in turn, is associated with the PCNA protein (P in the figure).

This protein acts as a "clothespin" that attaches the polymerase complex to the replicated DNA strand. It is believed that in the “buttoned” state, it wraps around the DNA chain like a ring (Fig.). This prevents premature dissociation of polymerases from this chain.

It is clear that DNA polymerases sequentially insert deoxyribonucleotides into the DNA chain being built - complementary to the nuclotides of the parent chain.

But, in addition, these enzymes apparently have a number of other important activities. However, for eukaryotic DNA polymerases, the distribution of these activities is not yet completely clear. Therefore, we present information on similar bacterial enzymes.

In bacteria, the main "work" for DNA replication is performed by DNA polymerase III, which has a dimer structure. It is with it that the “clamp” of the PCNA protein type is associated.

So, in addition to DNA polymerase activity, DNA polymerase III has one more - 3 "→ 5" exonuclease. The latter works in cases where a mistake is made and the “wrong” nucleotide is included in the chain being built. Then, having recognized the defect in base pairing, the enzyme cleaves off the last nucleotide from the growing (3 "-) end, after which it again begins to work as a DNA polymerase.

Thus, there is a constant control of the system over the result of its activities.

c) As we know, new DNA chains are first formed in the form of fragments - relatively short (Okazaki fragments) and very long. And each of them starts with a primer RNA.

When the enzyme complex moving along the parent strand reaches the RNA primer of the previous fragment, the "clamp" binding DNA polymerase III to the parent DNA strand opens and this enzyme stops working. DNA polymerase I comes into play (we are still talking about bacterial enzymes). It attaches to the 3"-end of the growing fragment (Fig. 1.14). In this case, the enzyme no longer has a stable connection with this fragment and with the parent chain, but instead it has not even two, but three activities.

The first of them is the “front” or 5'→3'-exonuclease activity: sequential cleavage of nucleotides from the 5' end of the RNA primer of the previous fragment.

The end of "own" fragment (DNA polymerase activity).

And, finally, like DNA polymerase III, it "does not forget" to check and, if necessary, correct its activity - with the help of "posterior", or 3'→5'-exonuclease activity directed at the fragment being extended.

The function of DNA polymerase I is exhausted when the growing fragment comes close to the deoxyribonucleotides of the previous fragment.

As far as eukaryotes are concerned, here the functional analog of bacterial DNA polymerase III is apparently a complex of α- and δ-DNA polymerases; while the corrective 3'→5'-exonuclease activity is inherent in δ-DNA polymerase.

The functions of DNA polymerase I are also distributed between two enzymes: 5" → 3" exonuclease activity (removal of the RNA primer) is probably carried out by a special nuclease (H in Fig. 1.11), and DNA polymerase activity (building "gaps") - DNA polymerase β (the one that is also involved in repair).

d) Speaking of polymerization enzymes, one cannot fail to mention the most difficult of the problems associated with them. It's about about the synthesis of the delayed strand of DNA: as we know, the direction of this synthesis is opposite general direction propagation of the replication fork.

There are at least two hypotheses explaining this contradiction.

According to one of them (Fig. 1.15, A), the enzyme complex periodically stops the formation of the leading strand, switches to the second parent strand, and synthesizes the next Okazaki fragment of the lagging strand. Then it returns to the first parent strand and continues to lengthen the leading strand of the DNA under construction.

According to another version (Fig. 1.15, B), a loop is formed on the second strand of parental DNA (the template of the lagging strand) during replication. Therefore, the direction of formation of the Okazaki fragment in the inner part of the loop begins to coincide with the direction of movement of the polymerase complex. Then the latter can almost simultaneously form both strands of DNA at once - both leading and lagging.

This may be related to the fact that bacterial DNA polymerase III is a dimer, while in eukaryotes α- and δ-DNA polymerases form a single complex. But even with such a mechanism, it is easy to see that the delayed chain cannot be formed continuously, but only in the form of fragments.

Enzymes that terminate DNA replication

As a result of the action of all previous enzymes, each iosynthesized chain turns out to consist of fragments that are closely adjacent to each other.

"Stitching" of adjacent fragments is carried out by DNA ligase (L in Fig. 1.11). Like DNA polymerases, this enzyme forms an internucleotide (phosphodiester) bond.

But if in the polymerase reaction one of the participants is free dNTP (deoxyribonucleoside triphosphate), then in the DNA ligase reaction both participants are terminal dNMP (deoxyribonucleoside monophosphates) as part of the “crosslinked” fragments.

For this reason, the energy of the reaction is different, and conjugated hydrolysis of the ATP molecule is required.

Note also that DNA ligase "crosslinks" only those single-stranded fragments that are part of double-stranded DNA.

But that's not all. The DNA molecule will not be completely replicated unless a special process of replication of its ends, or telomeric regions, occurs.

The enzyme telomerase plays a key role in this process.

Primases. DNA replication requires RNA primers. RNA primers are synthesized by primase (Fig. 29.3), encoded by the dnaG gene.

Figure 29.3 shows that the primase consists of three domains:

■ - N-terminal domain (110 amino acids), contains a DNA-binding motif - a zinc finger;

■ – core (central) domain (322 amino acids) contains a catalytic center;

■ – C-terminal domain (151 amino acids) interacting with dnaB.

Primers synthesized by the E. coli primase begin with the pppAG sequence at the 5' end and are approximately 10-12 nucleotides long. Primases differ both in structure and specificity of action.

DNA ligases catalyze the processes of reunification of fragments of DNA chains, participating in the formation of covalent bonds between the 5_-P - and 3_-OH groups of neighboring deoxyribonucleotides. These enzymes also use the energy of macroergic bonds formed during the hydrolysis of ATP.

DNA replication occurs in three stages: initiation, elongation and termination.

In bacteria, the initiation of DNA replication begins at a unique site on the chromosome, the replication point, oriC, from which replication proceeds bidirectionally to the termination point (terminus). As a result, two replication forks are formed that move in opposite directions, i.e., both strands replicate simultaneously.

initiator protein dnaA binds to repetitive binding sites on oriC, forming a specialized nucleoprotein structure. This leads to a local divergence of the AT-rich sequence oriC, which serves as a binding site for the replicative helicase (dnaB), and protein dnaC/

Further dnaB activated by removal dnaC, moves a certain distance in the direction 5_→3_ and interacts with the primase dnaG. Primase synthesizes short RNA primers for the DNA polymerase III holoenzyme. An intermediate complex consisting of at least five proteins is formed at the site of initiation. One of them is protein. dnaB- can move along DNA using the energy of ATP hydrolysis, and also serves as a signal for primase activation (Fig. 29.5).

Primase is a component of the primosome, which is made up of several different subunits. The primosome also contains a complex of proteins DNA and DnaС, which near the replication fork is periodically involved in the formation of a specific secondary DNA structure suitable for recognition by primase.

The initiation of DNA replication ends with the formation of a replication fork and the synthesis of an RNA primer on the leading DNA strand (Fig. 29.5) due to the formation of a replication complex (Fig. 29.6).

In the process of elongation, the daughter polynucleotide chains of DNA are built up. Each replication fork contains at least two DNA polymerase III molecules associated with several accessory proteins. The latter include DNA topoisomerases (gyrases), which unwind the tightly folded DNA double helix, and helicases, which unwind double-stranded DNA into two strands.

The leading strand of DNA replicates continuously in the same direction as the replication fork. The lagging strand is read in the opposite direction of the replication fork. Overcoming the antiparallelism of DNA strands during replication is possibly achieved by the formation of a looped structure (Fig. 29.7).

First, short fragments of a new DNA strand are synthesized on the lagging strand, the so-called Okazaki fragments, named after their discoverer. Each fragment begins with a short RNA primer (primer) necessary for the functioning of DNA polymerase. DNA polymerase III completes this primer to a DNA fragment with a length of 1000-2000 deoxynucleotide units.

Continuation. See No. 11, 12, 13, 14, 15/2005

Biology lessons in science classes

Advanced Planning, Grade 10

3. Connection of nucleotides in a chain

Nucleotides are linked together in a condensation reaction. In this case, an ester bond arises between the 3 "carbon atom of the sugar residue of one nucleotide and the phosphoric acid residue of the other. As a result, unbranched polynucleotide chains are formed. One end of the polynucleotide chain (called the 5" end) ends with a phosphoric acid molecule attached to 5 " -carbon atom, the other (it is called the 3 "end) - a hydrogen ion attached to the 3" carbon atom. A chain of consecutive nucleotides constitutes the primary structure of DNA.

Thus, the skeleton of the polynucleotide chain is carbohydrate-phosphate, since nucleotides are connected to each other by forming covalent bonds (phosphodiester bridges), in which the phosphate group forms a bridge between the C 3 atom of one sugar molecule and the C 5 atom of the next. Strong covalent bonds between nucleotides reduce the risk of "breakdowns" of nucleic acids.

If a polynucleotide formed by four types of nucleotides contains 1000 links, then the number of possible variants of its composition is 4 1000 (this is a figure with 6 thousand zeros). Therefore, only four types of nucleotides can provide a huge variety of nucleic acids and the information that they contain.

4. Formation of a double-stranded DNA molecule

In 1950, the English physicist Maurice Wilkins obtained an x-ray of DNA. She showed that the DNA molecule has a certain structure, the decoding of which would help to understand the mechanism of its functioning. X-rays obtained on highly purified DNA allowed Rosalind Franklin to see a clear cross-shaped pattern - the identification mark of the double helix. It became known that the nucleotides are located at a distance of 0.34 nm from each other, and there are 10 of them per turn of the helix.

The diameter of a DNA molecule is about 2 nm. From the radiographic data, however, it was not clear how the two strands were held together.

The picture became completely clear in 1953, when the American biochemist James Watson and the English physicist Francis Crick, having considered the totality of known data on the structure of DNA, came to the conclusion that the sugar-phosphate backbone is located on the periphery of the DNA molecule, and the purine and pyrimidine bases are in the middle.

D. Watson and F. Crick found that two DNA polynucleotide chains are twisted around each other and around a common axis. DNA chains are antiparallel (multidirectional), i.e. opposite the 3 "end of one chain is the 5" end of the other (imagine two snakes twisted into a spiral - the head of one to the tail of the other). The spiral is usually twisted to the right, but there are cases of the formation of a left spiral.

5. Chargaff rules. The essence of the principle of complementarity

Even before the discovery of Watson and Crick, in 1950 the Australian biochemist Edwin Chargaff established that 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 \u003d T, G \u003d C), or the total number of purine nitrogenous bases is equal to the total number of pyrimidine nitrogenous bases (A + G \u003d 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 complementary to each other (from lat. complementum- addition). We have already encountered several times the manifestation of complementarity (the active center of the enzyme and the substrate molecule are complementary to each other; antigen and antibody are complementary to each other).

Why is this principle followed? To answer this question, we need to remember chemical nature 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.

In this way, 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.

Formation of hydrogen bonds between complementary nitrogenous bases

There are two hydrogen bonds between adenine and thymine (A=T), and three hydrogen bonds between guanine and cytosine (G=C). Such a connection of nucleotides provides, firstly, the formation of the maximum number of hydrogen bonds, and secondly, the same distance between the chains along the entire length of the helix.

From all of the above, it follows that, knowing the sequence of nucleotides in one helix, you can find out the order of the nucleotides on the other helix.

The double complementary strand makes up the secondary structure of DNA. The helical shape of DNA is its tertiary structure.

III. Consolidation of knowledge

Generalizing conversation in the course of studying new material; problem solving.

Task 1. A section of one of the chains of the DNA molecule was studied in the laboratory. It turned out that it consists of 20 monomers, which are arranged in the following sequence: G-T-G-T-A-A-C-G-A-C-C-G-A-T-A-C-T-G -T-A.
What can be said about the structure of the corresponding section of the second strand of the same DNA molecule?

Knowing that the chains of a DNA molecule are complementary to each other, we determine the sequence of nucleotides of the second chain of the same DNA molecule: C-A-C-A-T-T-G-C-T-G-G-C-T-A-T- G-A-C-A-T.

Task 2. On a fragment of one DNA chain, the nucleotides are arranged in the sequence: A-A-G-T-C-T-A-C-G-T-A-T ...

1. Draw a diagram of the structure of the second strand of this DNA molecule.
2. What is the length in nm of this DNA fragment if one nucleotide is about 0.34 nm?
3. How many (in%) nucleotides are contained in this fragment of the DNA molecule?

1. We complete the second strand of this fragment of the DNA molecule, using the rule of complementarity: T-T-C-A-G-A-T-G-C-A-T-A.
2. Determine the length of this DNA fragment: 12x0.34=4.08 nm.
3. Calculate the percentage of nucleotides in this DNA fragment.

24 nucleotides - 100%
8A - x%, hence x = 33.3% (A);
because according to the Chargaff rule, A = T, which means the content of T = 33.3%;
24 nucleotides - 100%
4D - x%, hence x \u003d 16.7% (G);
because according to the Chargaff rule, G=C, which means that the content of C=16.6%.

Answer: T-T-C-A-G-A-T-G-C-A-T-A; 4.08 nm; A=T=33.3%; G=C=16.7%

Task 3. What will be the composition of the second DNA strand if the first contains 18% guanine, 30% adenine and 20% thymine?

1. Knowing that the chains of the DNA molecule are complementary to each other, we determine the content of nucleotides (in%) in the second chain:

because in the first chain G = 18%, then in the second chain C = 18%;
because in the first chain A=30%, so in the second chain T=30%;
because in the first chain T=20%, so in the second chain A=20%;

2. Determine the content in the first chain of cytosine (in%).

determine the proportion of cytosine in the first DNA strand: 100% - 68% = 32% (C);

if in the first chain C=32%, then in the second chain G=32%.

Answer: C=18%; T=30%; A=20%; G=32%

Task 4. In a DNA molecule, there are 23% of adenyl nucleotides from total number nucleotides. Determine the amount of thymidyl and cytosyl nucleotides.

1. According to the Chargaff rule, we find the content of thymidyl nucleotides in a given DNA molecule: A=T=23%.
2. Find the sum (in%) of the content of adenyl and thymidyl nucleotides in a given DNA molecule: 23% + 23% = 46%.
3. Find the sum (in%) of the content of guanyl and cytosyl nucleotides in this DNA molecule: 100% - 46% = 54%.
4. According to the Chargaff rule, in the DNA molecule G=C, in total they account for 54%, and individually: 54% : 2 = 27%.

Answer: T=23%; C=27%

Task 5. 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?

1. Determine how many adenyl nucleotides are in a given DNA molecule: 8625: 345 = 25.
2. According to the Chargaff rule, A=G, i.e. in this DNA molecule A=T=25.
3. Determine how much of the total molecular weight of this DNA is the share of guanyl nucleotides: 69,000 - (8625x2) = 51,750.
4. Determine the total number of guanyl and cytosyl nucleotides in this DNA: 51 750:345=150.
5. Determine the content of guanyl and cytosyl nucleotides separately: 150:2 = 75;
6. Determine the length of this DNA molecule: (25 + 75) x 0.34 = 34 nm.

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

Problem 6. 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 pairs of nucleotides are there in the DNA of one cell (1 nm = 10–6 mm)?

1. Convert centimeters to millimeters and nanometers: 102 cm = 1020 mm = 1,020,000,000 nm.
2. Knowing the length of one nucleotide (0.34 nm), we determine the number of base pairs contained in the DNA molecules of the human gamete: (10 2 x 10 7): 0.34 = 3 x 10 9 pairs.

Answer: 3x109 pairs.

IV. Homework

Study the paragraph of the textbook and the notes made in the class (content, molecular weight of nucleic acids, nucleotide structure, Chargaff's rule, the principle of complementarity, the formation of a double-stranded DNA molecule), solve problems after the text of the paragraph.

Lesson 16-17. Classes of cellular RNA and their functions. differences between DNA and RNA. DNA replication. mRNA synthesis

Equipment: tables on general biology; nucleotide structure scheme; model of the structure of DNA; diagrams and drawings illustrating the structure of RNA, the processes of replication and transcription.

I. Knowledge Test

Card work

Card 1. Indicate the fundamental differences in the structure of the DNA molecule from the molecules of other biopolymers (proteins, carbohydrates).

Card 2. What is the huge information capacity of DNA based on? For example, the DNA of mammals contains 4–6 billion bits of information, which corresponds to a library of 1.5–2 thousand volumes. How is this function reflected in the structure?

Card 3. When heated, DNA, like proteins, denatures. What do you think happens to the double helix?

Card 4. Fill in the gaps in the text: “Two strands of the DNA molecule are facing each other .... The chains are connected..., and against the nucleotide containing adenine there is always a nucleotide containing..., and against containing cytosine - containing.... This principle is called the principle ... . The order... in the molecule... for each organism... determines the sequence... in... . So DNA is... . DNA is localized mainly in ... cells in eukaryotes and in ... cells in prokaryotes.

Oral knowledge test on questions

1. Nucleic acids, their content in living matter, molecular weight.
2. NC - non-periodic polymers. The structure of the nucleotide, types of nucleotides.
3. Connection of nucleotides in a chain.
4. Formation of a double-stranded DNA molecule.
5. Chargaff rules. The essence of the principle of complementarity.

Validation problem solving given in the textbook.

II. Learning new material

1. RNA and its meaning

Proteins form the basis of life. Their functions in the cell are very diverse. However, proteins "can't" reproduce. And all the information about the structure of proteins is contained in genes (DNA).

At higher organisms proteins are synthesized in the cytoplasm of the cell, and DNA is hidden behind the shell of the nucleus. Therefore, DNA cannot directly serve as a template for protein synthesis. This role is performed by another nucleic acid - RNA.

The RNA molecule is an unbranched polynucleotide with a tertiary structure. It is formed by one polynucleotide chain, and although the complementary nucleotides included in it are also capable of forming hydrogen bonds between themselves, these bonds occur between the nucleotides of one chain. RNA chains are much shorter than DNA chains. If the content of DNA in a cell is relatively constant, then the content of RNA fluctuates greatly. The greatest amount of RNA in cells is observed during protein synthesis.

RNA belongs the main role in the transmission and implementation of hereditary information. In accordance with the function and structural features, several classes of cellular RNA are distinguished.

2. Classes of cellular RNA and their functions

There are three main classes of cellular RNA.

1. Informational (mRNA), or matrix (mRNA). Its molecules are the most diverse in terms of size, molecular weight (from 0.05x10 6 to 4x10 6) and stability. They make up about 2% of the total amount of RNA in the cell. All mRNAs are carriers of genetic information from the nucleus to the cytoplasm, to the site of protein synthesis. They serve as a matrix (working drawing) for the synthesis of a protein molecule, as they determine the amino acid sequence (primary structure) of a protein molecule.

2. Ribosomal RNA (rRNA). They make up 80–85% of the total RNA content in the cell. Ribosomal RNA consists of 3–5 thousand nucleotides. It is synthesized in the nucleoli of the nucleus. In complex with ribosomal proteins, rRNA forms ribosomes - organelles on which protein molecules are assembled. The main significance of rRNA is that it provides the initial binding of mRNA and ribosome and forms the active center of the ribosome, in which peptide bonds are formed between amino acids during the synthesis of the polypeptide chain.

3. Transfer RNAs(T RNA). tRNA molecules usually contain 75-86 nucleotides. Molecular mass tRNA molecules are about 25 thousand. tRNA molecules play the role of intermediaries in protein biosynthesis - they deliver amino acids to the site of protein synthesis, that is, to ribosomes. The cell contains more than 30 types of tRNA. Each type of tRNA has its own unique nucleotide sequence. However, all molecules have several intramolecular complementary regions, due to the presence of which all tRNAs have a tertiary structure resembling a clover leaf in shape.

3. Differences between DNA and RNA molecules

Filling in the table by students with subsequent verification.

Signs of comparison
Location in the cell	Nucleus, mitochondria, chloroplasts	Nucleus, ribosomes, centrioles, cytoplasm, mitochondria and chloroplasts
The structure of the macromolecule	Double unbranched linear polymer coiled	Single polynucleotide chain
Monomers	Deoxyribonucleotides	Ribonucleotides
Composition of nucleotides	Purine (adenine, guanine) and pyrimidine (thymine, cytosine) nitrogenous bases; deoxyribose (C5); phosphoric acid residue	Purine (adenine, guanine) and pyrimidine (uracil, cytosine) nitrogenous bases; ribose (C5); phosphoric acid residue
	Custodian of hereditary information	Intermediary in the implementation of genetic information

4. DNA replication

One of the unique properties of the DNA molecule is its ability to self-replicate - to reproduce exact copies of the original molecule. Due to this, the transfer of hereditary information from the mother cell to the daughter cells during division is carried out. The process of self-replication of a DNA molecule is called replication (reduplication).

Replication is a complex process involving enzymes (DNA polymerases). For replication, the DNA double helix must first be untwisted. This is also done by special enzymes - helicases breaking hydrogen bonds between bases. But untwisted areas are very sensitive to damaging factors. To keep them in an unprotected state for as little time as possible, synthesis on both chains occurs simultaneously.

But in maternal DNA, the two chains of the double helix are antiparallel - opposite the 3'-end of one strand is the 5'-end of the other, and the DNA polymerase enzyme can only "move" in one direction - from the 3'-end to the 5'-end of the template strand. . Therefore, replication of one half of the parent molecule, beginning with a 3'-nucleotide, is switched on after the double helix is ​​untwisted and is believed to be continuous. The replication of the second half of the molecule begins a little later and not from the beginning (where the 5'-nucleotide is located, which prevents the reaction), but at some distance from it. At the same time, DNA polymerase moves in the opposite direction, synthesizing a relatively short fragment. The structure that appears at this moment is called replication fork. As the double helix is ​​untwisted, the replication fork shifts: on the second chain, the synthesis of the next segment begins, going towards the beginning of the previous, already synthesized fragment. Then these separate fragments on the second matrix chain (they are called fragments of Okazaki) are linked together by the enzyme DNA ligase into a single strand.

Diagram of the structure of the DNA replication fork

During replication, the energy of ATP molecules is not consumed, since for the synthesis of daughter chains during replication, not deoxyribonucleotides are used (they contain one phosphoric acid residue), but deoxyribonucleoside triphosphates(contain three residues of phosphoric acid). When deoxyribonucleoside triphosphates are included in the polynucleotide chain, two terminal phosphates are cleaved off, and the released energy is used to form an ester bond between nucleotides.

As a result of replication, two double "daughter" helices are formed, each of which preserves (preserves) one of the halves of the original "maternal" DNA unchanged. The second chains of "daughter" molecules are synthesized from nucleotides anew. It got the name semi-conservative DNA.

5. Synthesis of RNA in the cell

The reading of RNA from a DNA template is called transcription(from lat. transcription- rewriting). It is carried out by a special enzyme - RNA polymerase. Three different RNA polymerases have been found in eukaryotic cells that synthesize different classes RNA.

Transcription is also an example of a template synthesis reaction. The RNA chain is very similar to the DNA chain: it also consists of nucleotides (ribonucleotides, very similar to deoxyribonucleotides). RNA is read from the DNA region in which it is encoded in accordance with the principle of complementarity: uracil RNA becomes opposite DNA adenine, cytosine opposite guanine, adenine opposite thymine, and guanine opposite cytosine.

Within a given gene, only one strand of two complementary DNA strands serves as a template for RNA synthesis. This circuit is called working.

In accordance with accepted conventions, the beginning of the gene is depicted on the left in the diagrams. The non-working (non-coding) strand of the DNA molecule in this case will have a 5 "end, while the working (coding) strand will have the opposite. The RNA polymerase enzyme attaches to promoter(a specific sequence of DNA nucleotides that the enzyme "recognizes" due to chemical affinity and which is located at the 3 "end of the corresponding section of the template DNA chain). Only by attaching to the promoter, RNA polymerase is able to start RNA synthesis from free ribonucleoside triphosphates present in the cell. Energy for RNA synthesis is contained in the macroenergetic bonds of ribonucleoside triphosphates.

III. Consolidation of knowledge

Generalizing conversation in the course of learning new material. The solution of the problem.

Task. The DNA molecule consists of two chains - the main one, on which mRNA is synthesized, and the complementary one. Write down the order of nucleotides in the synthesized mRNA, if the order of nucleotides in the main (working) DNA strand is as follows: C-G-C-T-G-A-T-A-G.

Using the principle of complementarity, we determine the order of nucleotides in the mRNA synthesized along the working DNA chain: G-C-G-A-C-U-A-U-C.

Answer: G-Ts-G-A-Ts-U-A-U-Ts

IV. Homework

Study the textbook paragraph (RNA, its main classes and functions, differences between DNA and RNA, replication and transcription).

Lesson 18

Equipment: general biology tables, nucleotide structure diagram, DNA structure model, diagrams and drawings illustrating the structure of RNA, replication and transcription processes.

I. Knowledge Test

Oral test of knowledge on questions.

1. RNA and its importance in the cell.
2. Classes of cellular RNA and their functions ( three students).
3. Replication, its mechanism and significance.
4. Transcription, its mechanism and significance.

Biological dictation "Comparison of DNA and RNA"

The teacher reads the theses under the numbers, the students write down in the notebook the numbers of those theses that are suitable in content to their version.

Option 1 - DNA; option 2 - RNA.

1. Single-stranded molecule.
2. Double-stranded molecule.
3. Contains adenine, uracil, guanine, cytosine.
4. Contains adenine, thymine, guanine, cytosine.
5. Ribose is a part of nucleotides.
6. Nucleotides contain deoxyribose.
7. Contained in the nucleus, chloroplasts, mitochondria, centrioles, ribosomes, cytoplasm.
8. Contained in the nucleus, chloroplasts, mitochondria.
9. Participates in the storage, reproduction and transmission of hereditary information.
10. Participates in the transfer of hereditary information.

Option 1 - 2; 4; 6; eight; 9;

Option 2 - 1; 3; 5; 7; 10.

Problem solving

Task 1. Chemical analysis showed that 28% of the total number of nucleotides of this mRNA is adenine, 6% is guanine, and 40% is uracil. What should be the nucleotide composition of the corresponding section of double-stranded DNA, the information from which is “rewritten” by this mRNA?

1. Knowing that the chain of the RNA molecule and the working chain of the DNA molecule are complementary to each other, we determine the content of nucleotides (in%) in the working chain of DNA:

in the mRNA chain G = 6%, which means that in the working DNA chain C = 6%;

in the mRNA chain A \u003d 28%, which means that in the working DNA chain T \u003d 28%;

in the mRNA chain Y \u003d 40%, which means that in the working DNA chain A \u003d 40%;

2. Determine the content of the mRNA chain (in%) of cytosine.

determine the proportion of cytosine in the mRNA chain: 100% - 74% = 26% (C);

if in the mRNA chain C=26%, then in the working DNA chain G=26%.

Answer: C=6%; T=28%; A=40%; G=26%

Task 2. On a fragment of one DNA chain, the nucleotides are located in the sequence: A-A-G-T-C-T-A-A-C-G-T-A-T. Draw a diagram of the structure of a double-stranded DNA molecule. What is the length of this DNA fragment? How many (in%) nucleotides are in this DNA strand?

1. By the principle of complementarity, it builds the second strand of a given DNA molecule: T-T-C-A-G-A-T-T-G-C-A-T-A.

2. Knowing the length of one nucleotide (0.34 nm), we determine the length of this DNA fragment (in DNA, the length of one chain is equal to the length of the entire molecule): 13x0.34 = 4.42 nm.

3. Calculate the percentage of nucleotides in this DNA chain:

13 nucleotides - 100%
5 A - x%, x \u003d 38% (A).
2 G - x%, x \u003d 15.5% (G).
4 T – x%, x=31% (T).
2 C - x%, x \u003d 15.5% (C).

Answer: T-T-C-A-G-A-T-T-G-C-A-T-A; 4.42 nm; A=38; T=31%; G=15.5%; C=15.5%.

Conducting independent work

Option 1

1. Fragments of one chain of the DNA molecule are given: C-A-A-A-T-T-G-G-A-C-G-G-G. Determine the content (in%) of each type of nucleotide and the length of this fragment of the DNA molecule.

2. 880 guanyl nucleotides were found in the DNA molecule, which make up 22% of the total number of nucleotides of this DNA? Determine how many other nucleotides are contained (individually) in this DNA molecule. What is the length of this DNA?

Option 2

1. Fragments of one chain of the DNA molecule are given: A-G-C-C-G-G-G-A-A-T-T-A. Determine the content (in%) of each type of nucleotide and the length of this fragment of the DNA molecule.

2. In the DNA molecule, 250 thymidyl nucleotides were found, which make up 22.5% of the total number of nucleotides of this DNA. Determine how many other nucleotides are contained (individually) in this DNA molecule. What is the length of this DNA?

IV. Homework

Repeat the material for the main classes organic matter found in living matter.

To be continued

DNA (right strand): GTA - ACC - TAT - CCG

DNA (left strand): CAT - TGG - ATA - GHZ

mRNA: GUA - ACC - UAU - CCG

Transcription

47. How many molecules of pentose - deoxyribose on a DNA segment, if information about a protein weighing 10,000 daltons is encoded on this gene segment. Mg (nucleotide) = 340; Mr (amino acids) = 100)

Number of amino acids in a protein = 10000/100 = 100

The number of nucleotides in a mature mRNA = 100 * 3 = 300 (since each amino acid is encoded by a triplet of nucleotides)

The number of nucleotides in the DNA gene = 300 * 2 = 600 (since DNA is double-stranded)

Number of deoxyribose residues in a DNA gene = number of nucleotides in a DNA gene = 600.

The number of nucleotides in DNA and, accordingly, the number of deoxyribose residues, determined based on the number of amino acids in a protein molecule, was produced without taking into account introns (non-coding fragments), but taking into account only coding regions (exons)

Answer: 600 deoxyribose residues.

In the Drosophila fly, the gray color of the body dominates over the black. When crossing gray flies, 1390 gray flies and 460 black flies appeared in the offspring. Make a scheme of inheritance and indicate the genotypes of parents and offspring

A - gray body color, a - black body color

F 1 1390 A_, 460 aa

gray black

Since the splitting in the offspring from crossing gray (with a dominant trait) individuals is close to 3: 1, then according to Mendel's second law (the law of trait splitting), the parents are heterozygotes.

Therefore, the inheritance pattern, genotypes of parents and offspring will be:

F 1 1AA, 2Aa, 1aa

gray black

Parents are heterozygous for the gene that determines body color (Aa), in the offspring there is a splitting according to genotype 1 (AA): 2 (Aa): 1 (aa), and according to phenotype 3 (A_, gray): 1 (aa, black).

In radishes, the root crop can be long, round and oval. When crossing plants with oval root crops, 121 plants with long root crops were obtained, 119 - with round ones, 243 - with oval ones. What can be the offspring during self-pollination of plants that have 1) a long root crop; 2) round root vegetable

Due to the fact that when crossing phenotypically identical plants (with an oval root crop) in the offspring, a splitting close to 1 (long root crop): 2 (oval root crop): 1 (round root crop) was obtained, then, firstly, the crossed parent plants according to Mendel's second law (the law of splitting of traits), they are heterozygous, and, secondly, the elongated form of the root crop does not completely dominate over the round one (incomplete dominance of the trait or the intermediate nature of inheritance), since splitting by phenotype corresponds to splitting by genotype. Due to the fact that 50% of individuals in the offspring had an oval root crop, heterozygous individuals are characterized by an oval root crop.

Let AA be an elongated root crop, Aa an oval root crop, and aa a round root crop.

Then the inheritance pattern when crossing individuals with an oval root crop will be as follows:

oval oval

F 1 1AA, 2Aa, 1aa

elongated oval round

1) With self-pollination of plants with a long root crop (AA), we get plants only with a long root crop:

long long

2) when self-pollinating plants with a round root crop (aa), we get plants with only a round root crop:

round round

50. What area of ​​the sea area (in m 2) is needed to feed one seal weighing 300 kg (water accounts for 60%) in the food chain: plankton - fish - seal. Bioproductivity of plankton is 600 g/m2

% dry residue in the body of a pike = 100-60 = 40%

m dry residue in the body of a pike = 300*40/100 = 120 kg

plankton ® fish ® seal

12000 kg 1200 kg 120 kg

Based on the productivity of plankton (0.6 kg / m 2), we determine the area of ​​​​the sea area necessary to feed the seal:

0.6 kg ® 1 m 2

120 kg ® x m 2

Field area \u003d 12000 / 0.6 \u003d 20000 m 2

Thus, to feed a pike, an area of ​​\u200b\u200bthe sea area is 20,000 m 2

A fragment of an mRNA molecule has the following nucleotide sequence: UGC-AAG-CUG-UUU-AUA. Determine the sequence of amino acids in a protein molecule. To do this, use the table of the genetic code

mRNA: UGC-AAG-CUG-UUU-AUA

peptide: cysteine ​​- lysine - leucine - phenylalanine - isoleucine

Broadcast

Answer: cysteine ​​- lysine - leucine - phenylalanine - isoleucine.

52. A mature mRNA molecule consists of 240 nucleotides. How many nucleotides are there in DNA, which was the template for the synthesis of this mRNA molecule, if introns account for 20%?

% exon nucleotides in immature mRNA = 100-20 = 80%

Number of nucleotides in immature mRNA = 240*100/80 = 300

The number of nucleotides in the DNA section from which this mRNA was copied = 300 * 2 = 600 (since DNA is double-stranded)

Exons are coding regions of genes, introns are non-coding polynucleotide sequences in genes, they can be longer than exons and presumably perform regulatory and structural functions. In the course of RNA maturation, non-coding regions copied from introns (processing) are cut out from it, and coding regions copied from exons are combined in the desired sequence (splicing).

Answer: the number of nucleotides in DNA = 600.

When crossing heterozygous red-fruited tomatoes with yellow-fruited ones, 352 plants with red fruits were obtained. The rest of the plants had yellow fruits. Determine how many plants had a yellow color? (the red color of the fruit is a dominant trait)

The red color of fruits in tomatoes is dominant. Let A be the red color of the fruit and let A be the yellow color of the fruit.

red yellow

red yellow

When a heterozygous individual is crossed with a recessive homozygote (analyzing cross), the split in F 1 is 1: 1 (50% heterozygotes that show a dominant trait and 50% recessive homozygotes that show a recessive trait). Consequently, there will be about the same number of yellow-flowered plants as red-fruited ones (i.e. 352 plants).

Answer: about 352 plants had a yellow color.

Enamel hypoplasia is inherited as an X-linked dominant trait, and six-fingeredness is inherited as an autosomal dominant trait. In a family where the mother is six-fingered, and the father has hypoplasia of tooth enamel, a five-fingered healthy boy was born. Indicate the genotypes of all family members and draw up a scheme of inheritance

Let X A - hypoplasia of tooth enamel, X a - normal enamel, B - six-fingered, b - five-fingered (normal)

Genotypes of parents and child: mother - X - X - Bb (six-fingered), father - X A Y_ _ (enamel hypoplasia), son - X a Ubb

P X - X - Bb x X A Y_ _

Six-toed enamel hypoplasia

five-fingered, normal enamel

Due to the fact that a five-fingered healthy boy was born to these parents, the genotypes of the mother and father will be as follows: X - X a Bb (mother), X A Y_ b (father).

Due to the fact that the condition of the problem does not say anything about the state of the enamel in the mother and the number of fingers in the father, there are 2 possible variants of the genotypes of the parents and, accordingly, 2 inheritance patterns:

1) P X a X a Bb x X A Ubb 2) P X A X a Bb x X A UVb

….. normal enamel, enamel hypoplasia enamel hypoplasia enamel hypoplasia

…….six-fingered five-fingered six-fingered……. six-fingered

F 1 X a Ubb F 1 X a Ubb

normal enamel normal enamel

five-fingered five-fingered

55. Determine the area of ​​the river, which is necessary in order to feed pike perch weighing 1 kg (40% dry matter). In the food chain: phytoplankton - herbivorous fish - zander. Phytoplankton productivity is 500 g/m2

% dry residue in the body of pike perch = 100-60 = 40%

m dry residue in the body of pike perch = 1*40/100 = 0.4 kg

According to the rule of Charles Elton's ecological pyramid, the total biomass of organisms, the energy contained in it, and the number of individuals decrease as one ascends from the lowest to the highest trophic level; at the same time, approximately 10% of the biomass and associated energy goes to each subsequent level. In this regard, the biomass of various links in the food chain will be:

phytoplankton ® herbivorous fish ® zander

40 kg 4 kg 0.4 kg

Based on the productivity of phytoplankton (0.5 kg / m 2), we determine the area of ​​\u200b\u200bthe sea area necessary to feed pike perch:

0.5 kg ® 1 m 2

40 kg ® x m 2

Field area \u003d 40 / 0.5 \u003d 80 m 2

Thus, to feed pike perch, an area of ​​\u200b\u200bthe sea area is 80 m 2

56. A section of a protein molecule has the following amino acid sequence: asparagine-isoleucine-proline-tryptophan-lysine. Determine one of the possible sequences of nucleotides in the DNA molecule (use the table of the genetic code)

peptide: asparagine-isoleucine-proline-tryptophan-lysine

mRNA: AAU - AUU - CCU - UGG - AAA

DNA (inf.thread): TTA - TAA - GGA - ACC - TTT

DNA (2nd strand): AAT - ATT - CCT - TGG - AAA

Transcription- the process of mRNA synthesis on a DNA matrix, is carried out according to principle of complementarity of nucleic polypeptides: nucleotide adenine is complementary (forms hydrogen bonds) to nucleotide thymine in DNA or nucleotide uracil in RNA, nucleotide cytosine is complementary to nucleotide guanine in DNA or RNA.

Broadcast- the process of protein synthesis on the mRNA matrix, is carried out on ribosomes with the participation of tRNA, each of which delivers a certain amino acid for protein synthesis. tRNA is a triplet of nucleotides (anticodon), which, according to the principle of complementarity, interacts with a specific triplet (codon) of mRNA.

A fragment of a DNA molecule recreated on the basis of a peptide and, accordingly, a mature mRNA molecule does not reflect the presence of introns (non-coding fragments), but includes only coding regions (exons).

The DNA molecule consists of 3600 nucleotides. Determine the number of complete helical turns in the given molecule. Determine the amount of t-RNA that will take part in the biosynthesis of the protein encoded in this gene

Number of base pairs in a DNA molecule = 3600/2 = 1800

The number of complete helical turns in a given DNA fragment = 1800/10 = 180 (because each turn of the DNA double helix includes 10 base pairs)

The number of nucleotides in one strand of DNA = 3600/2 = 1800 (since DNA is double-stranded)

The number of amino acids encoded in this DNA fragment (excluding the possible presence of introns in it) = 1800/3 = 600 (since each amino acid is encoded by a triplet of nucleotides)

The number of tRNA molecules involved in the biosynthesis of this protein = 600, since each amino acid is delivered by a specific tRNA molecule.

Transcription- the process of mRNA synthesis on a DNA matrix, is carried out according to principle of complementarity of nucleic polypeptides: nucleotide adenine is complementary (forms hydrogen bonds) to nucleotide thymine in DNA or nucleotide uracil in RNA, nucleotide cytosine is complementary to nucleotide guanine in DNA or RNA.

Broadcast- the process of protein synthesis on the mRNA matrix, is carried out on ribosomes with the participation of tRNA, each of which delivers a certain amino acid for protein synthesis. tRNA is a triplet of nucleotides (anticodon), which, according to the principle of complementarity, interacts with a specific triplet (codon) of mRNA.

When solving this problem, it was not possible to take into account the possible presence of intron (non-coding) regions in the DNA molecule, as a result of which the obtained amount of amino acids of the protein encoded in this DNA fragment, and, accordingly, the amount of tRNA required for the synthesis of this protein, may be overestimated.

Answer: the number of complete turns in a DNA molecule = 180; amount of tRNA = 600.

As a result of crossing two animals with wavy hair, 20 offspring were obtained, 15 of them with wavy six, and 5 with smooth hair. How many of the offspring are heterozygotes? Write an inheritance scheme

Due to the fact that when crossing phenotypically identical animals with each other in F1, a splitting of 3: 1 was obtained (15 animals with wavy hair and 5 with smooth hair), then according to the second law of Mendel (or the law of splitting of characters), the crossed parents were heterozygous and wavy wool dominates over smooth. Let A be wavy wool and let A be smooth wool.

Inheritance scheme:

wavy wavy

G A, a ....... A, a

F 1 AA, 2Aa, aa

wavy, smooth

% of heterozygous offspring = 50% of the total number of offspring or 2/3 of individuals with wavy hair, the number of heterozygous offspring = 15 * 2/3 = 10.

In butterflies, the female gender is determined by the XY chromosomes, and the male gender by the XX chromosomes. The trait "cocoon color" is sex-linked. White color cocoon is a dominant trait. What will be the offspring from crossing a white-coated female with a dark-coated male?

Let X A be a white cocoon, then X a is a dark cocoon

P X A Y x X a X a

white cocoon dark cocoon

female male

G X A, Y X a

F 1 X A X a, X a Y

white cocoon dark cocoon

male female

All males in F1 will have a white cocoon and all females will have a dark cocoon. In general, the splitting without taking into account gender will be 1:1.

60. Based on the rule of the ecological pyramid, determine what area of ​​the biocenosis will feed an owl weighing 2 kg in the grain-mice-owl food chain. The number of mice and the number of owls. Productivity of plant biocenosis 400 g/m 2

m dry residue in the body of an owl = 2 kg

According to the rule of Charles Elton's ecological pyramid, the total biomass of organisms, the energy contained in it, and the number of individuals decrease as one ascends from the lowest to the highest trophic level; at the same time, approximately 10% of the biomass and associated energy goes to each subsequent level. In this regard, the biomass of various links in the food chain will be:

grain ® mice ® ​​owl

200 kg 20 kg 2 kg

Based on the productivity of the biocenosis (0.4 kg / m 2), we determine the area of ​​\u200b\u200bthe biocenosis necessary to feed the owl:

0.4 kg ® 1 m 2

200 kg ® x m 2

Field area \u003d 200 / 0.4 \u003d 500 m 2

Thus, to feed an owl, an area of ​​\u200b\u200bthe biocenosis is 500 m 2