For this task you can get 3 points on the Unified State Exam in 2020

The topic of assignment 28 of the Unified State Exam in biology was “Supraorganismal systems and the evolution of the world.” Many schoolchildren note the difficulty of this test due to the large volume it covers. educational material, and also due to the construction of the ticket. In task No. 28, the compiler - the Russian FIPI, the Federal Institute of Pedagogical Measurements, offers six answer options for each question, of which any number from one to all six can be correct. Sometimes the question itself contains a hint - how many options you will have to choose (“Which three features out of the six listed are characteristic of animal cells”), but in most cases the student must decide for himself the number of answer options he chooses as correct.

Questions on task 28 of the Unified State Exam in biology may also touch on the basics of biology. Be sure to repeat before the exams - what is the absence of artificial and natural ecosystems, aquatic and terrestrial, meadow and field, what does the rule of the ecological pyramid sound like and where it is applicable, what is biogeocenosis and agrocenosis. Some questions are logical in nature; you need not only to rely on the theory from the school textbook, but also to think logically: “In a mixed forest, plants are arranged in tiers, and this is the reason for the decrease in competition between the birch and another living organism. Which one?” The following answers include cockchafer, bird cherry, mushrooms, rose hips, hazel, and mice. In this case, the student must remember that competition is always for the same resources, in this case (with a tiered arrangement of plants) - for light, so from the list you need to select only trees and shrubs - bird cherry, rose hips and hazel.

The assignment relates to highest level difficulties. For the correct answer you will receive 3 points.

It takes approximately up to 10-20 minutes.

To complete task 28 in biology you need to know:

• how to (make crossbreeding schemes), ecology, evolution;

Tasks for training

Task No. 1

The hamster color gene is linked to the X chromosome. The X A genome is determined by the brown color, the X B genome is black. Heterozygotes have a tortoiseshell coloration. Five black hamsters were born from a female tortoiseshell and a black male. Determine the genotypes of parents and offspring, as well as the nature of inheritance of traits.

Task No. 2

In fruit flies, black body color dominates over gray, and normal wings dominate over curved ones. Two black flies with normal wings are crossed. The offspring of F 1 are phenotypically uniform - with a black body and normal wings. What are the possible genotypes of the crossed individuals and offspring?

Task No. 3

Humans have four phenotypes according to blood groups: I(0), II(A), III(B), IV(AB). The gene that determines the blood group has three alleles: I A, I B, i 0, and the i 0 allele is recessive with respect to the IA and IB alleles. The color blindness gene d is linked to the X chromosome. A woman with blood type II (heterozygote) and a man with blood group III (homozygote) entered into marriage. It is known that the woman’s father suffered from color blindness, her mother was healthy. The man's relatives never had this disease. Determine the genotypes of the parents. Indicate possible genotypes and phenotypes (blood group number) of children. Make a diagram for solving the problem. Determine the probability of having color-blind children and children with blood group II.

Task No. 4

In corn, the genes for brown color and smooth seed shape are dominant over the genes for white color and wrinkled shape.

When plants with brown smooth seeds were crossed with plants with white seeds and wrinkled seeds, 4006 brown smooth seeds and 3990 white wrinkled seeds were obtained, as well as 289 white smooth and 316 brown wrinkled corn seeds. Make a diagram for solving the problem. Determine the genotypes of the parent corn plants and their offspring. Justify the appearance of two groups of individuals with characteristics different from their parents.

Using the pedigree shown in the figure, determine and explain the nature of inheritance of the trait (dominant or recessive, sex-linked or not) highlighted in black. Determine the genotypes of the descendants indicated in the diagram by numbers 3, 4, 8, 11 and explain the formation of their genotypes.

Explanation.

The trait highlighted in black is recessive, linked to the X chromosome: X a,

because there is a “leap” through a generation. A man with a sign (8) has a daughter without a sign (11), and grandchildren - one with a sign (12), the second without (13), that is, they receive a Y chromosome from their father (10), and from their mother (11) one X a, the other X A.

Genotypes of people indicated in the diagram by numbers 3, 4, 8, 11:

3 - female carrier - X A X a

4 - man without sign - X A Y

8 - a man with the sign - X and Y

11 - female carrier - X A X a

Source: Unified State Examination in Biology 05/30/2013. Main wave. Far East. Option 4.

Elena Ivanova 11.04.2016 12:36

Please explain why the genotype of the first woman (without number) is HAHA, because she could also be a carrier?

Natalia Evgenievna Bashtannik

Maybe. This is a "guess" based on offspring. Because it is not important for us to solve, we can write both options on the diagram, or even like this: X A X -

Nikita Kaminsky 11.06.2016 23:28

Why can’t there just be a recessive gene that is not sex-linked?

Then the parents in the first generation are homozygous (father aa, mother AA), children 1, 2, 3 are heterozygous Aa, men 4 and 5 are also carriers of Aa, children 7 and 8 in the second generation have the trait, and 6 is a carrier. In the third generation, Father and Mother are again homozygous, daughter 11 and her husband 10 are heterozygous, and they have two sons, one with the trait, the other without, possibly a carrier.

Natalia Evgenievna Bashtannik

maybe, but there is a greater probability that there is clutch, less “?”, and based on the rules for solving these problems.

A mother and father who are phenotypically without the trait give birth to a SON with the trait; it can be assumed that the trait is linked to the X chromosome.

Tobias Rosen 09.05.2017 18:26

The solution is not entirely correct.

This diagram contains an alternative solution - containing fewer assumptions:

In fact, all we can say based on the problem data is a list of what we can exclude. We can exclude dominant linkage to X, we can exclude linkage to Y, we can exclude AA x aa in the cross itself, we can exclude that the trait is provided by a dominant allele.

We cannot exclude recessive linkage with X and we cannot exclude autosomal recessive inheritance - there is not enough data for this and an insufficient number of descendants and crosses.

To ignore the small number of crossings and descendants is to assume that the law of large numbers must also apply to small numbers. Which is complete nonsense. Should not. On the contrary: the statistical fact is that the smaller the sample, the greater the expected deviation from the “correct split.”

Natalia Evgenievna Bashtannik

If a problem can be solved in two ways, then it is better to write both. If the criteria include a decision that the trait is linked to the X chromosome: X a, then they may not give a full point.

Among the tasks on genetics on the Unified State Exam in biology, 6 main types can be distinguished. The first two - to determine the number of gamete types and monohybrid crossing - are most often found in part A of the exam (questions A7, A8 and A30).

Problems of types 3, 4 and 5 are devoted to dihybrid crossing, inheritance of blood groups and sex-linked traits. Such tasks make up the majority of C6 questions in the Unified State Exam.

The sixth type of task is mixed. They consider the inheritance of two pairs of traits: one pair is linked to the X chromosome (or determines human blood groups), and the genes of the second pair of traits are located on autosomes. This class of tasks is considered the most difficult for applicants.

This article outlines theoretical foundations of genetics necessary for successful preparation for task C6, as well as solutions to problems of all types are considered and examples are given for independent work.

Basic terms of genetics

Gene- this is a section of a DNA molecule that carries information about the primary structure of one protein. A gene is a structural and functional unit of heredity.

Allelic genes (alleles)- different variants of one gene, encoding an alternative manifestation of the same trait. Alternative signs are signs that cannot be present in the body at the same time.

Homozygous organism- an organism that does not split according to one or another characteristic. Its allelic genes equally influence the development of this trait.

Heterozygous organism- an organism that produces cleavage according to certain characteristics. Its allelic genes have different effects on the development of this trait.

Dominant gene is responsible for the development of a trait that manifests itself in a heterozygous organism.

Recessive gene is responsible for a trait whose development is suppressed by a dominant gene. A recessive trait occurs in a homozygous organism containing two recessive genes.

Genotype- a set of genes in the diploid set of an organism. The set of genes in a haploid set of chromosomes is called genome.

Phenotype- the totality of all the characteristics of an organism.

G. Mendel's laws

Mendel's first law - the law of hybrid uniformity

This law was derived based on the results of monohybrid crosses. For the experiments, two varieties of peas were taken, differing from each other in one pair of characteristics - the color of the seeds: one variety was yellow in color, the second was green. The crossed plants were homozygous.

To record the results of crossing, Mendel proposed the following scheme:

Yellow color of seeds
- green color of seeds

(parents)
(gametes)
(first generation)	(all plants had yellow seeds)

Statement of the law: when crossing organisms that differ in one pair of alternative characteristics, the first generation is uniform in phenotype and genotype.

Mendel's second law - the law of segregation

Plants were grown from seeds obtained by crossing a homozygous plant with yellow colored seeds with a plant with green colored seeds and obtained by self-pollination.

	(plants have a dominant trait - recessive)

Statement of the law: in the offspring obtained from crossing first-generation hybrids, there is a split in phenotype in the ratio , and in genotype -.

Mendel's third law - the law of independent inheritance

This law was derived from data obtained from dihybrid crosses. Mendel considered the inheritance of two pairs of characteristics in peas: color and seed shape.

As parental forms, Mendel used plants homozygous for both pairs of traits: one variety had yellow seeds with smooth skin, the other had green and wrinkled seeds.

Yellow color of seeds, - green color of seeds,
- smooth form, - wrinkled form.

	(yellow smooth).

Mendel then grew plants from seeds and obtained second-generation hybrids through self-pollination.

	The Punnett grid is used to record and determine genotypes Gametes

There was a split into phenotypic classes in the ratio. All seeds had both dominant traits (yellow and smooth), - the first dominant and second recessive (yellow and wrinkled), - the first recessive and second dominant (green and smooth), - both recessive traits (green and wrinkled).

When analyzing the inheritance of each pair of traits, the following results are obtained. In parts of yellow seeds and parts of green seeds, i.e. ratio . Exactly the same ratio will be for the second pair of characteristics (seed shape).

Statement of the law: when crossing organisms that differ from each other in two or more pairs of alternative traits, genes and their corresponding traits are inherited independently of each other and combined in all possible combinations.

Mendel's third law is true only if the genes are located in different pairs of homologous chromosomes.

Law (hypothesis) of “purity” of gametes

When analyzing the characteristics of hybrids of the first and second generations, Mendel established that the recessive gene does not disappear and does not mix with the dominant one. Both genes are expressed, which is only possible if the hybrids form two types of gametes: some carry a dominant gene, others carry a recessive one. This phenomenon is called the gamete purity hypothesis: each gamete carries only one gene from each allelic pair. The hypothesis of gamete purity was proven after studying the processes occurring in meiosis.

The hypothesis of the "purity" of gametes is the cytological basis of Mendel's first and second laws. With its help, it is possible to explain the splitting by phenotype and genotype.

Analysis cross

This method was proposed by Mendel to determine the genotypes of organisms with a dominant trait that have the same phenotype. To do this, they were crossed with homozygous recessive forms.

If, as a result of crossing, the entire generation turned out to be the same and similar to the analyzed organism, then one could conclude: the original organism is homozygous for the trait being studied.

If, as a result of crossing, a split in the ratio was observed in a generation, then the original organism contains genes in a heterozygous state.

Inheritance of blood groups (AB0 system)

Inheritance of blood groups in this system is an example of multiple allelism (the existence of more than two alleles of one gene in a species). In the human population, there are three genes encoding red blood cell antigen proteins that determine people's blood types. The genotype of each person contains only two genes that determine his blood type: group one; second and ; third and fourth.

Inheritance of sex-linked traits

In most organisms, sex is determined during fertilization and depends on the number of chromosomes. This method is called chromosomal sex determination. Organisms with this type of sex determination have autosomes and sex chromosomes - and.

In mammals (including humans), the female sex has a set of sex chromosomes, while the male sex has a set of sex chromosomes. The female sex is called homogametic (forms one type of gametes); and the male one is heterogametic (forms two types of gametes). In birds and butterflies, the homogametic sex is male, and the heterogametic sex is female.

The Unified State Exam includes tasks only for traits linked to the - chromosome. They mainly concern two human characteristics: blood clotting (- normal; - hemophilia), color vision (- normal, - color blindness). Tasks on the inheritance of sex-linked traits in birds are much less common.

In humans, the female sex can be homozygous or heterozygous for these genes. Let's consider possible genetic sets in a woman using hemophilia as an example (a similar picture is observed with color blindness): - healthy; - healthy, but is a carrier; - sick. The male sex is homozygous for these genes, because -chromosome does not have alleles of these genes: - healthy; - is ill. Therefore, most often men suffer from these diseases, and women are their carriers.

Typical USE tasks in genetics

Determination of the number of gamete types

The number of gamete types is determined using the formula: , where is the number of gene pairs in the heterozygous state. For example, an organism with a genotype does not have genes in a heterozygous state, i.e. , therefore, and it forms one type of gametes. An organism with a genotype has one pair of genes in a heterozygous state, i.e. , therefore, and it forms two types of gametes. An organism with a genotype has three pairs of genes in a heterozygous state, i.e. , therefore, and it forms eight types of gametes.

Mono- and dihybrid crossing problems

For monohybrid crossing

Task: Crossed white rabbits with black rabbits (black color is the dominant trait). In white and black. Determine the genotypes of parents and offspring.

Solution: Since segregation according to the studied trait is observed in the offspring, therefore, the parent with the dominant trait is heterozygous.

	(black)	(white)
	(black) : (white)

For dihybrid crossing

Dominant genes are known

Task: Crossed normal-sized tomatoes with red fruits with dwarf tomatoes with red fruits. All plants were of normal growth; - with red fruits and - with yellow ones. Determine the genotypes of parents and offspring if it is known that in tomatoes, red fruit color dominates yellow, and normal growth dominates dwarfism.

Solution: Let us designate dominant and recessive genes: - normal growth, - dwarfism; - red fruits, - yellow fruits.

Let's analyze the inheritance of each trait separately. All descendants have normal growth, i.e. no segregation for this trait is observed, therefore the initial forms are homozygous. Segregation is observed in fruit color, so the original forms are heterozygous.

		(dwarfs, red fruits)
	(normal growth, red fruits) (normal growth, red fruits) (normal growth, red fruits) (normal growth, yellow fruits)

Dominant genes unknown

Task: Two varieties of phlox were crossed: one has red saucer-shaped flowers, the second has red funnel-shaped flowers. The offspring produced were red saucer, red funnel, white saucer and white funnel. Determine the dominant genes and genotypes of the parental forms, as well as their descendants.

Solution: Let's analyze the splitting for each characteristic separately. Among the descendants of plants with red flowers are, with white flowers -, i.e. . That's why it's red, - White color, and the parental forms are heterozygous for this trait (since there is segregation in the offspring).

There is also a split in flower shape: half of the offspring have saucer-shaped flowers, the other half have funnel-shaped flowers. Based on these data, it is not possible to unambiguously determine the dominant trait. Therefore, we accept that - saucer-shaped flowers, - funnel-shaped flowers.

	(red flowers, saucer-shaped)	(red flowers, funnel-shaped)
	Gametes

Red saucer-shaped flowers,
- red funnel-shaped flowers,
- white saucer-shaped flowers,
- white funnel-shaped flowers.

Solving problems on blood groups (AB0 system)

Task: the mother has the second blood group (she is heterozygous), the father has the fourth. What blood types are possible in children?

Solution:

	(the probability of having a child with the second blood group is , with the third - , with the fourth - ).

Solving problems on the inheritance of sex-linked traits

Such tasks may well appear in both Part A and Part C of the Unified State Examination.

Task: a carrier of hemophilia married a healthy man. What kind of children can be born?

Solution:

	girl, healthy () girl, healthy, carrier () boy, healthy () boy with hemophilia ()

Solving problems of mixed type

Task: A man with brown eyes and a blood type married a woman with brown eyes and a blood type. They had a blue-eyed child with a blood type. Determine the genotypes of all individuals indicated in the problem.

Solution: Brown eye color dominates blue, therefore - brown eyes, - Blue eyes. The child has blue eyes, so his father and mother are heterozygous for this trait. The third blood group can have a genotype or, the first - only. Since the child has the first blood group, therefore, he received the gene from both his father and mother, therefore his father has the genotype.

	(father)	(mother)
	(was born)

Task: A man is colorblind, right-handed (his mother was left-handed) married to a woman with normal vision (her father and mother were completely healthy), left-handed. What kind of children can this couple have?

Solution: In a person, better control of the right hand dominates over left-handedness, therefore - right-handed, - left-handed. The genotype of the man (since he received the gene from a left-handed mother), and women - .

A colorblind man has the genotype, and his wife has the genotype, because. her parents were completely healthy.

R
	right-handed girl, healthy, carrier () left-handed girl, healthy, carrier () right-handed boy, healthy () left-handed boy, healthy ()

Problems to solve independently

1. Determine the number of gamete types in an organism with genotype.
2. Determine the number of gamete types in an organism with genotype.
3. Crossed tall plants with short plants. B - all plants are medium in size. What will it be?
4. Crossed a white rabbit with a black rabbit. All rabbits are black. What will it be?
5. Two rabbits with gray fur were crossed. In with black wool, - with gray and with white. Determine the genotypes and explain this segregation.
6. A black hornless bull was crossed with a white horned cow. We got black hornless, black horned, white horned and white hornless. Explain this split if black color and lack of horns are dominant characteristics.
7. Drosophila flies with red eyes and normal wings were crossed with fruit flies with white eyes and defective wings. The offspring are all flies with red eyes and defective wings. What will be the offspring from crossing these flies with both parents?
8. A blue-eyed brunette married a brown-eyed blonde. What kind of children can be born if both parents are heterozygous?
9. A right-handed man with a positive Rh factor married a left-handed woman with a negative Rh factor. What kind of children can be born if a man is heterozygous only for the second characteristic?
10. The mother and father have the same blood type (both parents are heterozygous). What blood type is possible in children?
11. The mother has a blood type, the child has a blood type. What blood type is impossible for a father?
12. The father has the first blood group, the mother has the second. What is the probability of having a child with the first blood group?
13. A blue-eyed woman with a blood type (her parents had a third blood group) married a brown-eyed man with a blood type (his father had blue eyes and a first blood group). What kind of children can be born?
14. A hemophilic man, right-handed (his mother was left-handed) married a left-handed woman with normal blood (her father and mother were healthy). What children can be born from this marriage?
15. Strawberry plants with red fruits and long-petioled leaves were crossed with strawberry plants with white fruits and short-petioled leaves. What kind of offspring can there be if red color and short-petioled leaves dominate, while both parent plants are heterozygous?
16. A man with brown eyes and a blood type married a woman with brown eyes and a blood type. They had a blue-eyed child with a blood type. Determine the genotypes of all individuals indicated in the problem.
17. Melons with white oval fruits were crossed with plants that had white spherical fruits. The offspring produced the following plants: with white oval, white spherical, yellow oval and yellow spherical fruits. Determine the genotypes of the original plants and descendants, if the white color of the melon is dominant over the yellow one, oval shape fruit - above the spherical.

Answers

1. type of gametes.
2. types of gametes.
3. type of gametes.
4. high, medium and low (incomplete dominance).
5. black and white.
6. - black, - white, - gray. Incomplete dominance.
7. Bull: , cow - . Offspring: (black hornless), (black horned), (white horned), (white hornless).
8. - Red eyes, - white eyes; - defective wings, - normal. Initial forms - and, offspring.
   Crossing results:
   A)
9. - Brown eyes, - blue; - dark hair, - blond. Father mother - .
   

   	- brown eyes, dark hair - brown eyes, blond hair - blue eyes, dark hair - blue eyes, blond hair

10. - right-handed, - left-handed; - Rh positive, - Rh negative. Father mother - . Children: (right-handed, Rh positive) and (right-handed, Rh negative).
11. Father and mother - . Children may have a third blood group (probability of birth - ) or first blood group (probability of birth - ).
12. Mother, child; he received the gene from his mother, and from his father - . The following blood groups are impossible for the father: second, third, first, fourth.
13. A child with the first blood group can only be born if his mother is heterozygous. In this case, the probability of birth is .
14. - Brown eyes, - blue. Female Male . Children: (brown eyes, fourth group), (brown eyes, third group), (blue eyes, fourth group), (blue eyes, third group).
15. - right-handed, - left-handed. Man Woman . Children (healthy boy, right-handed), (healthy girl, carrier, right-handed), (healthy boy, left-handed), (healthy girl, carrier, left-handed).
16. - red fruits, - white; - short-petioled, - long-petioled.
    Parents: and. Offspring: (red fruits, short-petioled), (red fruits, long-petioled), (white fruits, short-petioled), (white fruits, long-petioled).
    Strawberry plants with red fruits and long-petioled leaves were crossed with strawberry plants with white fruits and short-petioled leaves. What kind of offspring can there be if red color and short-petioled leaves dominate, while both parent plants are heterozygous?
17. - Brown eyes, - blue. Female Male . Child:
18. - white color, - yellow; - oval fruits, - round. Source plants: and. Offspring:
    with white oval fruits,
    with white spherical fruits,
    with yellow oval fruits,
    with yellow spherical fruits.

Average general education

Line UMK V.V. Pasechnik. Biology (10-11) (basic)

Line UMK Ponomareva. Biology (10-11) (B)

Biology

Unified State Examination in Biology 2018: task 27, basic level

Experience shows that it is easier to get a high Unified State Exam score in biology if you solve problems as carefully as possible basic level. In addition, compared to last year, even basic tasks have become somewhat more complicated: they require a more complete, common answer. The solution will come to the student if he thinks a little, gives explanations, and gives arguments.
Together with an expert, we analyze examples of typical problems of line No. 27, clarify the solution algorithm, and consider different variants of the problems.

Task 27: what's new?

Some of the tasks in this line have changed: now it is more often necessary to predict the consequences of a mutation in a gene section. First of all, you will encounter variants of tasks on gene mutations, but it is also appropriate to repeat chromosomal and genomic mutations.

In general, task number 27 this year is presented with very diverse options. Some of the tasks are related to protein synthesis. It is important to understand here: the algorithm for solving a problem depends on how it is formulated. If the task begins with the words “it is known that all types of RNA are transcribed into DNA,” this is one synthesis sequence, but it may be proposed to simply synthesize a fragment of a polypeptide. Regardless of the wording, it is extremely important to remind students how to write DNA nucleotide sequences correctly: without spaces, hyphens or commas, and in a continuous sequence of characters.

To solve problems correctly, you need to carefully read the question, paying attention to additional comments. The question may sound like this: what changes can occur in a gene as a result of mutation if one amino acid in a protein is replaced by another? What property of the genetic code determines the possibility of the existence of different fragments of a mutated DNA molecule? The task may also be given to restore a DNA fragment in accordance with the mutation.

If the problem contains the wording “explain, using your knowledge about the properties of the genetic code,” it would be appropriate to list all the properties that are known to students: redundancy, degeneracy, non-overlapping, etc.

What topics must be studied in order to successfully solve line 27 problems?

• Mitosis, meiosis, development cycles of plants: algae, mosses, ferns, gymnosperms, angiosperms.


• Microsporogenesis and macrosporogenesis in gymnosperms and angiosperms.

New things are being brought to the attention of students and teachers. tutorial, which will help you successfully prepare for the unified state exam in biology. The collection contains questions selected according to sections and topics tested on the Unified State Exam, and includes tasks different types and difficulty levels. Answers to all tasks are provided at the end of the manual. The proposed thematic assignments will help the teacher organize preparation for the unified state exam, and students will independently test their knowledge and readiness to take the final exam. The book is addressed to students, teachers and methodologists.

How to prepare?

1. Show students the diagrams and algorithms: how spores and gametes are formed in plants, how gametes and somatic cells are formed in animals. It is useful to ask students to model the patterns of mitosis and meiosis themselves: this helps to understand why haploid cells formed by meiosis later become diploid.


2. Turn on your visual memory. It is useful to remember illustrations of basic patterns of the evolution of the life cycle of various plants - for example, the cycle of alternation of generations in algae, ferns, and bryophytes. Unexpectedly, for some reason, questions related to the life cycle of pine often cause difficulties. The topic itself is not complicated: it is enough to know about microsporangia and megasporangia that they are formed by meiosis. It is necessary to understand that the cone itself is diploid: this is obvious to the teacher, but not always to the student.


3. Pay attention to the nuances of the wording. When describing some issues, it is necessary to make clarifications: in the life cycle of brown algae, there is an alternation of haploid gametophyte and diploid sporaphyte with a predominance of the latter (this way we will get rid of possible quibbles). A nuance in the topic of the life cycle of ferns: when explaining what and how spores are formed, one can answer in different ways. One option is from sporagon cells, and the other, more convenient, is from mother spore cells. Both answers are satisfactory.

Let's look at examples of problems

Example 1. The DNA chain fragment has the following sequence: TTTTGCGATGCCCCGCA. Determine the sequence of amino acids in the polypeptide and justify your answer. What changes can occur in the gene as a result of mutation in, if the third amino acid in the protein is replaced by the CIS amino acid? What property of the genetic code determines the possibility of the existence of different fragments of a mutated DNA molecule? Explain your answer using the genetic code table.

Solution. This problem can be easily broken down into elements that will form the correct answer. It is best to follow a proven algorithm:

1. determine the sequence of amino acids in the fragment;


2. we write what will happen when replacing one amino acid;


3. we conclude that there is degeneracy of the genetic code: one amino acid is encoded by more than one triplet (here you need the skill of solving such problems).

Example 2. The chromosome set of somatic wheat cells is 28. Determine the chromosome set and the number of DNA molecules in one of the ovule cells before the onset of meiosis, in anaphase of meiosis I and anaphase of meiosis II. Explain what processes occur during these periods and how they affect changes in the number of DNA and chromosomes.

Solution. Before us is a classic, well-known problem in cytology. It is important to remember here: if the problem asks you to determine the chromosome set and the number of DNA molecules, and also shows numbers, do not limit yourself to the formula: be sure to indicate the numbers.

The following steps are required in the solution:

1. indicate the initial number of DNA molecules. In this case it is 56 - since they double, but the number of chromosomes does not change;


2. describe anaphase of meiosis I: homologous chromosomes diverge to the poles;


3. describe anaphase of meiosis II: the number of DNA molecules is 28, chromosomes are 28, sister chromatids-chromosomes diverge to the poles, since after the reduction division of meiosis I the number of chromosomes and DNA decreased by 2 times.

In this formulation, the answer is most likely to bring the desired high score.

Example 3. What chromosome set is characteristic of pine pollen grain and sperm cells? From what initial cells and as a result of what division are these cells formed?

Solution. The problem is formulated transparently, the answer is simple and easily broken down into components:

1. pollen grain and sperm cells have a haploid set of chromosomes;


2. pollen grain cells develop from haploid spores - by mitosis;


3. sperm - from pollen grain cells (generative cells), also by mitosis.

Example 4. Cattle have 60 chromosomes in their somatic cells. Determine the number of chromosomes and DNA molecules in ovarian cells in interphase before the start of division and after meiosis I division. Explain how such a number of chromosomes and DNA molecules is formed.

Solution. The problem is solved using the previously described algorithm. In interphase before the start of division, the number of DNA molecules is 120, chromosomes - 60; after meiosis I, 60 and 30, respectively. It is important to note in the answer that before division begins, DNA molecules double, but the number of chromosomes does not change; we are dealing with reduction division, so the number of DNA is reduced by 2 times.

Example 5. What chromosome set is characteristic of fern germ cells and gametes? Explain from what initial cells and as a result of what division these cells are formed.

Solution. This is the same type of problem where the answer can easily be broken down into three elements:

1. indicate the set of chromosomes of the germ n, gametes - n;


2. We must indicate that the prothallus develops from a haploid spore by mitosis, and gametes develop on the haploid prothallus by mitosis;


3. since the exact number of chromosomes is not indicated, you can limit yourself to the formula and write simply n.

Example 6. Chimpanzees have 48 chromosomes in their somatic cells. Determine the chromosome set and the number of DNA molecules in cells before the onset of meiosis, in anaphase of meiosis I and in prophase of meiosis II. Explain your answer.

Solution. As you may have noticed, in such tasks the number of answer criteria is precisely visible. In this case, they are: determine the set of chromosomes; define it in certain phases - and be sure to give explanations. It is most logical to give an explanation after each numerical answer in the answer. For example:

1. We give the formula: before the start of meiosis, the set of chromosomes and DNA is 2n4c; at the end of interphase, DNA doubling occurred, the chromosomes became bichromatid; 48 chromosomes and 96 DNA molecules;


2. in anaphase of meiosis, the number of chromosomes does not change and is equal to 2n4c;


3. Haploid cells that have a set of bichromatid chromosomes with the n2c set enter meiosis II prophase. Therefore, at this stage we have 24 chromosomes and 48 DNA molecules.

A new textbook is offered to students and teachers that will help them successfully prepare for the unified state exam in biology. The reference book contains all the theoretical material on the biology course necessary for passing the Unified State Exam. It includes all elements of content, verified by test materials, and helps to generalize and systematize knowledge and skills for a secondary (high) school course. Theoretical material presented in a concise, accessible form. Each section is accompanied by examples test tasks, allowing you to test your knowledge and degree of preparedness for the certification exam. Practical tasks correspond to the Unified State Exam format. At the end of the manual, answers to tests are provided that will help schoolchildren and applicants test themselves and fill in existing gaps. The manual is addressed to schoolchildren, applicants and teachers.

You can learn anything, but it is more important to learn to reflect and apply the learned knowledge. Otherwise, you will not be able to score adequate passing scores. During the educational process, pay attention to the formation of biological thinking, teach students to use language adequate to the subject, and work with terminology. There is no point in using a term in a textbook paragraph if it will not work in the next two years.