Among the tasks in genetics on the exam in biology, 6 main types can be distinguished. The first two - for determining the number of gamete types and monohybrid crossing - are found most often 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 most of the C6 questions in the exam.

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

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

Basic terms of genetics

Gene 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 inheritance.

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

Homozygous organism- an organism that does not split in one way or another. Its allelic genes equally affect the development of this trait.

Heterozygous organism- an organism that splits according to one or another characteristic. Its allelic genes affect the development of this trait in different ways.

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

Recessive gene is responsible for the trait, the development of which is suppressed by the dominant gene. A recessive trait is manifested in a homozygous organism containing two recessive genes.

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

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

G. Mendel's laws

Mendel's first law - the law of uniformity of hybrids

This law is derived from the results of monohybrid crossing. For the experiments, two varieties of peas were taken, differing from each other by one pair of signs - the color of the seeds: one variety had a yellow color, the second - green. The crosses were homozygous.

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

Yellow seed color
- green color of seeds

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

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

Mendel's second law - the law of splitting

From seeds obtained by crossing a homozygous plant with a yellow seed color with a plant with a green seed color, plants were grown and obtained by self-pollination.

The wording of the law: in the offspring obtained from crossing the first generation hybrids, there is a splitting according to the phenotype in the ratio, and according to the genotype -.

Mendel's third law - the law of independent inheritance

This law was deduced on the basis of data obtained from dihybrid crossing. Mendel considered the inheritance of two pairs of traits in peas: color and shape of seeds.

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

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

Then Mendel grew plants from seeds and, by self-pollination, obtained second-generation hybrids.

In there was a splitting into phenotypic class in the ratio. of all seeds had both dominant traits (yellow and smooth), - the first dominant and the second recessive (yellow and wrinkled), - the first recessive and the 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 characters (seed shape).

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

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

The law (hypothesis) of "purity" of gametes

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

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

Analyzing cross

This method was proposed by Mendel to elucidate the genotypes of organisms with a dominant trait that have the same phenotype. For 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 it could be concluded that the original organism is homozygous for the studied trait.

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

Inheritance of blood groups (AB0 system)

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

Inheritance of sex-linked traits

In most organisms, sex is determined during fertilization and depends on the set 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, the male sex. The female sex is called homogametic (forms one type of gametes); and the male is heterogametic (forms two types of gametes). In birds and butterflies, males are homogametic, and females are heterogametic.

The USE includes tasks only for characters linked to the -chromosome. Basically, they relate to two signs of a person: blood clotting (- norm; - hemophilia), color vision (- norm, - color blindness). Much less common is the problem of inheriting sex-linked traits in birds.

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

Typical USE tasks in genetics

Determination of the number of gamete types

Determination of the number of gamete types is carried out according to the formula:, where is the number of pairs of genes in a heterozygous state. For example, an organism with a genotype has no 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, it also 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.

Problems for mono- and dihybrid crossing

For monohybrid crossing

Task: White rabbits were crossed with black rabbits (black is dominant). In whites and blacks. Determine the genotypes of the parents and offspring.

Solution: Since in the offspring there is a splitting according to the studied trait, therefore, the parent with the dominant trait is heterozygous.

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

For dihybrid crossing

Dominant genes are known

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

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

Let's analyze the inheritance of each trait separately. All offspring are of normal height, i.e. splitting for this trait is not observed, therefore the original forms are homozygous. By the color of the fruit, splitting is observed, therefore the original forms are heterozygous.

Dominant genes unknown

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

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

Splitting is also observed in the shape of the flower: half of the offspring have saucer-shaped flowers, half are funnel-shaped. Based on these data, it is not possible to unambiguously determine the dominant feature. Therefore, we will assume that - saucer-shaped flowers, - funnel-shaped flowers.

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

Solving problems for 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:

Solving problems on the inheritance of sex-linked traits

Such tasks may well be encountered both in part A and in part C of the exam.

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

Solution:

Mixed problem solving

Task: A man with brown eyes and blood type married a woman with brown eyes and blood type. They had a blue-eyed baby with a blood group. Determine the genotypes of all persons indicated in the task.

Solution: Brown eyes dominate over 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 a genotype.

Task: The man is color blind, 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, the best control of the right hand dominates over the left-handedness, therefore - right-handed, - left-handed. Male genotype (since he received the gene from a left-handed mother), and women -.

A color-blind man has a genotype, and his wife has a genotype. her parents were completely healthy.

Tasks for independent solution

1. Determine the number of gamete types in an organism with a genotype.
2. Determine the number of gamete types in an organism with a genotype.
3. Crossed tall plants with low plants... B - all medium-sized plants. What will happen?
4. We crossed a white rabbit with a black rabbit. All rabbits are black. What will happen?
5. We crossed two rabbits with gray hair. B with black wool, - with gray and white. Identify genotypes and explain this splitting.
6. They crossed a black hornless bull with a white horned cow. We got black hornless, black horned, white horned and white hornless. Explain this cleavage if black color and lack of horns are dominant.
7. Drosophila with red eyes and normal wings were crossed with fruit flies with white eyes and defective wings. In the offspring, 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 rhesus factor. What kind of children can be born if a man is heterozygous only for the second trait?
10. The mother and father have a blood type (both parents are heterozygous). What blood group is possible in children?
11. The mother has a blood group, the child has a blood group. 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 baby with the first blood group?
13. A blue-eyed woman with a blood group (her parents had a third blood group) married a brown-eyed man with a blood group (his father had blue eyes and a first blood group). What kind of children can be born?
14. A right-handed hemophilic man (his mother was left-handed) married a left-handed woman with normal blood (her father and mother were healthy). What kind of children can be born from this marriage?
15. We crossed strawberry plants with red fruits and long-stemmed leaves with strawberry plants with white fruits and short-stemmed leaves. What offspring can there be if the red color and short-petiolate leaves are dominant, 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 baby with a blood group. Determine the genotypes of all persons indicated in the task.
17. Melons with white oval fruits were crossed with plants that had white globular fruits. In the offspring, the following plants were obtained: with white oval, white globular, yellow oval and yellow globular fruits. Determine the genotypes of the original plants and offspring, if the white color of the melon dominates over the yellow, the oval shape of the fruit - over 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, - light. Father mother - .
   

   	- brown eyes, dark hair - brown eyes, blonde hair - blue eyes, dark hair - blue eyes, blonde hair

10. - right-handed, - left-handed; - Rh positive, - 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; from his mother he received the gene, 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 be born only 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 petiolate, - long petiolate.
    Parents: and. Offspring: (red fruits, short petiolate), (red fruits, long petiolate), (white fruits, short petiolate), (white fruits, long petiolate).
    We crossed strawberry plants with red fruits and long-stemmed leaves with strawberry plants with white fruits and short-stemmed leaves. What offspring can there be if the red color and short-petiolate leaves are dominant, 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 globular fruits,
    with yellow oval fruits,
    with yellow globular fruits.

Secondary general education

Biology

Preparing for the exam in biology: text with errors

• Do not “train” students for specific assignments. Future surgeons, veterinarians, psychologists and other serious professionals must demonstrate a deep knowledge of the subject.
• Go beyond the tutorials. On profile exam graduates will have to demonstrate more than knowledge of the program.
• Use proven tutorials. With a wide variety of biology materials, many teachers choose publications of the Russian Textbook corporation.
• Allow variability in your answers. There is no need to present the reference formulation as the only correct one. The answer can be given in other words, contain additional information, differ from the standard in the form and sequence of presentation.
• Practice answering questions in writing. Pupils often do not know how to give complete written answers, even with a high level of knowledge.
• Get used to working with pictures. Some students do not know how to extract information from pictures for assignments.

• Demonstrate knowledge of terminology. This is especially important in the second part of the exam. Appeal with concepts (preferably literary).
• Express your thoughts clearly. Answers must be accurate and meaningful.
• Read the assignments carefully, consider all the criteria. If it says “Explain the answer”, “Provide evidence”, “Explain the meaning”, then for the absence of an explanation, the points are reduced.
• Write the correct definition. In task number 24, an error is not considered corrected if the answer contains only negative judgment.
• Proceed with the elimination method. In task number 24, first look for sentences that contain or do not exactly contain errors.

Examples of assignments number 24 and possible difficulties

Exercise: Find three errors in the above text. Indicate the numbers of sentences in which mistakes were made, correct them. Give the correct wording.

Example 1

Example 2

(1) Eukaryotic cells begin to prepare for division in prophase. (2) During this preparation, the process of protein biosynthesis takes place, DNA molecules are doubled, ATP is synthesized. (3) In the first phase of mitosis, the centrioles of the cell center, mitochondria and plastids are doubled. (4) Mitotic division consists of four phases. (5) In metaphase, the chromosomes line up in the equatorial plane. (6) Then, in anaphase, homologous chromosomes diverge to the poles of the cell. (7) The biological significance of mitosis lies in the fact that it ensures the constancy of the number of chromosomes in all cells of the body.

Response elements:(1) Preparation for fission begins in the interphase. (3) Doubling of all named organelles occurs in interphase. (6) Sister chromatids rather than homologous chromosomes diverge to the poles of the cell in mitosis.

Note: The student can write “chromatids-chromosomes”. The textbooks contain the phrase: "Chromatids are also chromosomes", so such a wording will not be considered an error or will become a reason for appeal if a score is lowered for it.

A new textbook is offered to the attention of students and teachers that will help them successfully prepare for the unified state exam in biology. The handbook contains all the theoretical material on the biology course required for passing the exam... It includes all the elements of the content, verified by control and measuring materials, and helps to generalize and systematize knowledge and skills for the course of secondary (full) school. The theoretical material is presented in a concise, accessible form. Each section is accompanied by examples of test items to test your knowledge and degree of preparation for the certification exam. Practical tasks correspond to format of the exam... At the end of the manual, you will find answers to tests that will help schoolchildren and applicants test themselves and fill in the gaps. The manual is addressed to schoolchildren, applicants and teachers.

Example 3

(1) Chromosomes contained in one cell of an animal are always paired, i.e. the same, or homologous. (2) Chromosomes of different pairs in organisms of the same species are also the same in size, shape, locations of primary and secondary constrictions. (3) The set of chromosomes contained in one nucleus is called the chromosome set (karyotype). (4) In any animal organism, somatic and germ cells are distinguished. (5) The nuclei of somatic and germ cells contain a haploid set of chromosomes. (6) Somatic cells are formed as a result of meiotic division. (7) Sex cells are required for the formation of a zygote.

Response elements:(2) Chromosomes of different pairs differ from each other in all of the above characteristics. (5) Somatic cells contain a diploid set of chromosomes. (6) Somatic cells are formed as a result of mitosis.

Note: Chromosomes are not always paired, so the student may identify the first sentence as erroneous. If he corrects the other three sentences correctly, the score will not be deducted.

Example 4

(1) Amphibians are vertebrates living in water and on land. (2) They swim well; swimming membranes are developed between the toes of tailless amphibians. (3) On land, amphibians move with the help of two pairs of five-toed limbs. (4) Amphibians breathe with the help of the lungs and skin. (5) Adult amphibians have a two-chambered heart. (6) Fertilization in tailless amphibians is internal; tadpoles develop from fertilized eggs. (7) Amphibians include lake frog, gray toad, water snake, crested newt.

Response elements:(5) The heart in tadpoles is two-chambered. (6) In the vast majority of tailless amphibians, fertilization is external. (7) The water snake is classified as a reptile.

Note: The limbs of frogs are correctly called five-toed, but the student can write that one pair of limbs in frogs is four-toed. Without the rest of the corrections provided, this paragraph will be considered erroneous.

A new textbook is offered to the attention of students and teachers that will help them successfully prepare for the unified state exam in biology. The collection contains questions, selected according to sections and topics, tested on the exam, and includes tasks of different types and levels of difficulty. At the end of the manual you will find answers to all tasks. The proposed thematic assignments will help the teacher organize preparation for the unified state exam, and the students independently test their knowledge and readiness for the final exam. The book is addressed to students, teachers and methodologists.

Examples of assignments number 25 and possible difficulties

Questions need to be answered.

Example 1

What are the formations on the roots of a legume plant? What type of relationships between organisms is established in these formations? Explain the significance of this relationship for both organisms.

Response elements: 1. Formations on the roots of leguminous plants are nodules containing nodule azobacteria. 2. The type of relationship of symbiosis of nitrogen-fixing bacteria and plants. 3. Nodule bacteria feed on organic matter of plants (plants provide bacteria with organic matter) 4. Nodule bacteria fix atmospheric nitrogen and provide.

Note: The student can be confused by the text of the assignment. Are you talking about the relationship between the organisms inhabiting the formations or between the plant and the organisms? Are there two or more organisms? Of course, the authors of the work strive for maximum clarity in the assignments, but inaccurate formulations are still encountered, and the graduate must be ready for this.

Example 2

How does a pine seed differ in structure from a fern spore? Please indicate at least three differences

Response elements: 1. The seed is a multicellular formation, the spore is unicellular. 2. The seed has a supply of nutrients, the spore does not have this supply. 3. There is an embryo in a seed, but does not have an embryo spore.

Note: The spore is not a plant germ. Pupils often confuse the concepts of "dispute" and "embryo" - this should be taken into account when preparing.

Example 3

List the membranes of the human eyeball and what functions they perform.

Response elements: 1. The tunica albuginea (sclera) - protection of internal structures; its transparent part - the cornea - protection and light refraction (optical function). 2. Choroid - blood supply to the eye (pigment layer - light absorption); its part - the iris - regulation of the light flux. 3. Retina - perception of light (or color) and transformation into nerve impulses (receptor function).

Note: This is a simple exercise in which students make many of the same mistakes. The guys do not write about the fact that the tunica albuginea passes into the cornea, do not write about the functions of the cornea associated with refraction, about the transition of the choroid into the iris, about the fact that the iris provides eye pigmentation. But students often mistakenly state that the lens and vitreous are also membranes of the eye.

Example 4

Where are the sympathetic nuclei of the autonomic nervous system located? In what cases is it activated and how does it affect the work of the heart?

Response elements: 1. The bodies of the first nuclei (neurons) lie in the central nervous system in the spinal cord. 2. The bodies of the second neurons lie on both sides along the spine. 3. VNS is activated in a state of strong excitement with vigorous activity of the body. 4. Strengthens the heart rate.

Note: Nervous system issues are always complex. It is worth carefully studying the options for tasks on this topic, as well as repeating the structure of the autonomic nervous system, its reflex arcs, the functions of the sympathetic and parasympathetic nervous systems.

In conclusion, we note that a graduate will pass the USE in biology for a high score only if there is motivation, diligence and hard work. The responsibility for preparing for the exam lies largely with the student himself. The teacher's task is to guide and, if possible, teach to learn.

The exam in biology is a selective exam and it will be taken by those who are confident in their knowledge. The Unified State Exam in Biology is considered a difficult subject, as the knowledge accumulated over the years of study is tested.

USE assignments in biology, different types were selected, for their solution you need confident knowledge of the main topics of the school biology course. On the basis of the teachers developed over 10 test tasks for each topic.

For topics that need to be studied when completing assignments, see from FIPI. For each task, its own algorithm of actions is prescribed, which will help in solving problems.

Changes in the KIM USE 2019 in biology:

• The model of the assignment in line 2 has been changed. Instead of the assignment with multiple choice for 2 points, an assignment for working with a table for 1 point is included.
• Maximum primary score decreased by 1 and amounted to 58 points.

The structure of the tasks for the exam in biology:

• Part 1- these are tasks from 1 to 21 with a short answer, about 5 minutes are given to complete.

Advice: read the wording of the questions carefully.

• Part 2- these are tasks from 22 to 28 with a detailed answer, approximately 10-20 minutes are allotted to complete.

Advice: express your thoughts in literature, answer the question in detail and comprehensively, define biological terms, even if this is not required in the assignments. The answer should have a plan, not write in solid text, but highlight points.

What is required of a student for the exam?

• Ability to work with graphic information (diagrams, graphs, tables) - its analysis and use;
• Multiple choice;
• Establishing compliance;
• Sequencing.

Points for each task in biology of the exam

In order to get the highest grade in biology, you need to score 58 primary points, which will be translated into one hundred on a scale.

• 1 point - for 1, 2, 3, 6 tasks.
• 2 points - 4, 5, 7-22.
• 3 points - 23-28.

How to Prepare for Biology Test Items

1. Repetition of the theory.
2. Correct allocation of time for each task.
3. Solving practical problems several times.
4. Checking the level of knowledge by solving tests online.

Register, study and get a high score!

Tasks from the Unified State Exam Bank in Biology with a Free Answer

1. The biological oxidation of organic substances in the human body is similar in chemical process to the combustion of fuel (coal, peat, wood). What products are common with combustion as a result of these processes? Compare the energetics of biological oxidation and combustion processes. What is their difference?

1) as a result of the oxidation of organic substances by oxygen, as in combustion, carbon dioxide and water are formed;

2) during combustion, all energy is released in the form of heat, and during biological oxidation, part of the energy is stored in ATP molecules

2. Why, according to the ecological pyramid rule, is there a decrease in energy in the terrestrial food chain from link to link?

1) the energy contained in organic substances at each link of the food chain is spent on vital processes;

2) some of the energy is dissipated in the form of heat.

3. Why does the nasal cavity need to be moist and clean for normal odor perception? Explain the answer.

1) the cavity must be moist, since the olfactory cells (receptors) are irritated only by substances dissolved in the mucus of the nasal cavity;

2) the abundant secretion of mucus prevents the access of substances to olfactory receptors

4. Build a food chain using all of the listed species: cruciferous fleas, polecat, snake, turnip leaves, frog. Identify the consumer of the second order in the chain and explain your choice.

1) turnip leaves → cruciferous fleas → frog → snake → polecat;

2) the consumer of the II order is a frog, as it feeds on the consumers of the I order

Response elements:

1) wet seeds will begin to germinate, while they breathe intensively and generate a lot of heat;

2) strong heating of a large number of seeds leads to the death of both germinated and non-germinated seeds

6. What are the formations on the roots of the depicted plant? What type of relationships between organisms does the figure illustrate? Explain the meaning

these relationships for both organisms.

Response elements:

1) formations on the roots of a legume plant are nodules containing nodule bacteria;

2) the type of mutually beneficial relationship - the symbiosis of bacteria (nitrogen-fixing bacteria) and a legume plant;

3) nodule bacteria feed on organic matter of plants;

4) nodule bacteria fix atmospheric nitrogen and provide legumes with nitrogen compounds

7. Using the picture, identify the isolation method that led to the appearance of the three related subspecies of the great tit and explain its consequences. What result of evolution can their reproductive isolation lead to?

Response elements:

1) geographical isolation led to the emergence of three subspecies of the great tit;

2) as a result of geographical isolation, crossing and exchange of genes between individuals of different populations stops,
each population forms its own gene pool;

3) reproductive isolation can lead to the formation of three related species of tits

8. The figure shows a diagram of speciation according to Charles Darwin. What an evolutionary process

leads to the formation of the new species shown in Figure III? What driving forces (factors) of evolution underlie this process? What form of natural selection takes place in this case?

Response elements:

1) divergence (divergence) of signs;

2) divergence is due to hereditary variability, the struggle for existence and natural selection;

3) the driving (disruptive) form of natural selection

8. Name the departments of the visual analyzer indicated in Figure 1 and 2. What is the function of each of these departments?

Response elements:

1) 1 - peripheral section (or retina, or receptors);

2) 2 - conduction department (or optic nerve);

3) the retina perceives and converts light irritation
into nerve impulses;

4) the optic nerve transmits a nerve impulse to the brain

9. Name the animal shown in the picture and indicate its type. Which organ systems are indicated by numbers 1 and 2? What functions do they perform?

Response elements:

1) the lancelet is depicted; type Chordates;

2) 1 – nervous system- participates in the nervous regulation of all body functions and the relationship with environment;

3) 2 – digestive system(intestine) - carries out the digestion of food and absorption of nutrients

10. Find three mistakes in the above text. Indicate the numbers of sentences in which mistakes were made, correct them.

1. Fungi and bacteria are classified as prokaryotes. 2. There is a wide variety of fungi: yeast, mold, cap fungi, etc. 3. A common feature of multicellular fungi is the formation of a vegetative body from thin branching filaments that form a mycelium. 4. The fungus cell has a cell wall composed of chitin and membrane organelles. 5. The reserve nutrient is glycogen. 6. Mushrooms have an autotrophic type of nutrition. 7. The growth of fungi stops after the maturation of the spores.

Response elements: mistakes were made in sentences:

1) 1 - mushrooms are eukaryotes;

2) 6 - fungi have a heterotrophic type of nutrition;

3) 7 - mushrooms grow throughout life

Instructions

To solve genetic problems, certain types of research are used. The method of hybridological analysis was developed by G. Mendel. It allows you to identify the patterns of inheritance of individual traits during sexual reproduction. The essence of this method is simple: when analyzing certain alternative traits, they are traced in the offspring. Also, an accurate record of the manifestation of each alternative trait and each individual individual of the offspring is carried out.

The main patterns of inheritance were also developed by Mendel. The scientist deduced three laws. Subsequently, they are so - Mendel's laws. The first is the law of uniformity of the first hybrids. Take two heterozygous individuals. When crossed, they will give two types of gametes. The offspring of these will appear in a ratio of 1: 2: 1.

Mendel's second law is the law of splitting. it is based on the fact that the dominant gene does not always suppress the recessive one. In this case, not all individuals among the first generation reproduce the characteristics of their parents - the so-called intermediate nature of inheritance appears. For example, when crossing homozygous with red flowers (AA) and white flowers (aa), offspring with pink ones is obtained. Incomplete dominance is fairly common. It is also found in some biochemical characters.

The third and last law is the law of independent combination of features. For the manifestation of this law, several conditions must be met: there must be no lethal genes, dominance must be complete, genes must be in different chromosomes.

The tasks of gender genetics stand apart. There are two types of sex chromosomes: the X chromosome (female) and the Y chromosome (male). Sex with two identical sex chromosomes is called homogametic. Sex determined by different chromosomes is called heterogametic. The sex of the future individual is determined at the time of fertilization. In the sex chromosomes, in addition to genes that carry information about the sex, there are others that have nothing to do with this. For example, the gene responsible for blood clotting is carried by the female X chromosome. Sex-linked traits are transmitted from the mother to the sons and daughters, and from the father only to the daughters.

Related Videos

Sources:

• solving problems in biology genetics
• for dihybrid crossing and for inheritance of traits

All tasks in genetics, as a rule, are reduced to several main types: calculated, to find out the genotype and to find out how the trait is inherited. Such tasks can be schematic or illustrated. However, for the successful solution of any problem, including a genetic one, it is necessary to carefully read its condition. The very same decision is based on the implementation of a number of specific actions.

You will need

• - notebook;
• - textbook on genetics;
• - a pen.

Instructions

First, you need to determine the type of the proposed task. To do this, you will have to find out how many gene pairs for the development of the proposed traits, what traits are considered. Find out homo- or heterozygous in this case, interbreed with each other, as well as whether the inheritance of a particular trait is associated with sex chromosomes.

Find out which of the features proposed for study is (weak), and which is dominant (strong). At the same time, when solving a genetic problem, it is necessary to start from the premise that the dominant trait in the offspring will always manifest itself phenotypically.

Determine the number and type of gametes (reproductive). It should be borne in mind that gametes can only be haploid. Accordingly, the distribution of chromosomes during their division occurs evenly: each of the gametes will contain only one chromosome taken from a homologous pair. As a result, the offspring receives a "half" set of chromosomes from each of its own.

Make a schematic record of the condition of the genetic problem in a notebook. In this case, the dominant traits for the homozygous subject are in the form of a combination of AA, for the heterozygous - Aa. A_ has an indeterminate genotype. The recessive feature is written as a combination of aa.

Analyze the results obtained and write down this numerical ratio. This will be the answer to the genetic task.

Related Videos

Helpful advice

In many such problems, the genotype of individuals proposed for crossing is not specified. That is why it is so important to be able to independently determine the genotype of parents by the phenotype or genotype of their offspring.

In the study of genetics, much attention is paid to problems, the solution of which must be found using the laws of gene inheritance. For most students of natural sciences, solving problems by genetics seems to be one of the most difficult things in biology. Nevertheless, it is found using a simple algorithm.

You will need

• - textbook.

Instructions

First, read the problem carefully and write down the schematic condition using special characters. Indicate what genotypes the parents have and what phenotype corresponds to them. Write down which children came out in the first and second generations.

Note which gene is dominant and which is recessive, if present in the condition. If splitting is given in the problem, also indicate it in the schematic record. For simple tasks, sometimes it is enough to write down a condition to understand the solution tasks.

To successfully solve the problem, you need to understand to which section it is: monohybrid, dihybrid or polyhybrid crossing, sex-linked inheritance, or the trait is inherited by genes. To do this, calculate what kind of splitting of the genotype or phenotype is observed in the offspring in the first generation. The condition can indicate the exact number of individuals with each genotype or phenotype, or the percentage of each genotype (phenotype) from. These data must be reduced to simple ones.

Pay attention to whether the offspring do not differ in traits depending on gender.

Each type of crossing is characterized by its own special splitting and phenotype. All this data is contained in the textbook, and it will be convenient for you to write these formulas on a separate sheet and use them when solving problems.

Now that you have discovered splitting, according to the principle of which hereditary traits are transmitted in your task, you can find out the genotypes and phenotypes of all individuals in the offspring, as well as the genotypes and phenotypes of the parents who participated in the crossing.

Everything tasks on biology are divided into tasks on molecular biology and tasks on genetics. In molecular biology there are several topics that include tasks: proteins, nucleic acids, DNA code and energy metabolism.

Instructions

Decide tasks on the topic "Proteins" using the following formula: m (min) = a / b * 100%, where m (min) is the molecular weight, a is the atomic or molecular weight of the component, b is the percentage of the component. The average molecular weight of one acidic residue is 120.

Calculate the necessary values ​​on the topic "Nucleic acids", adhering to Chargaff: 1. The amount of adenine is equal to the amount of thymine, and guanine - cytosine;
2. The number of purine bases is equal to the number of pyrimidine bases, i.e. A + G = T + C. In the chain of a DNA molecule, the distance between nucleotides is 0.34 nm. The relative molecular weight of one nucleotide is 345.

Solve problems on the topic "DNA Code" using a special table of genetic codes. Thanks to her, you will find out which acid this or that genetic code encodes.

Calculate the answer you need for problems on the topic "Energy exchange" by the reaction equation. One of the most common is: С6Н12О6 + 6О2 → 6СО2 + 6Н2О.

Find genetics using a special algorithm. First, identify which genes are dominant (A, B) and which are recessive (a, b). A dominant gene is a gene whose trait is manifested both in a homozygous state (AA, aa) and in a heterozygous state (Aa, Bb). A gene is called recessive, a trait of which is manifested only when the same genes meet, i.e. in a homozygous state. For example, yellow-seeded peas were crossed with seeded peas. The resulting pea plants were all yellow. It's obvious that yellow is the dominant feature. Write down the solution to this tasks so: A - the gene responsible for the yellow color of the seeds, and - the gene responsible for green color seed. R: AA x aa
G: A, a
F1: Aa Exist tasks of this type with several features, then designate one feature A or a, and the second B or b.

The study of genetics is accompanied by problem solving. They clearly show the operation of the law of gene inheritance. For most students, these tasks seem incredibly difficult. But, knowing the solution algorithm, you can easily cope with them.

Instructions

Two main types can be distinguished. In the first type of problem, the genotypes of the parents are known. It is necessary to determine the genotypes of the offspring. First, determine which allele is dominant. Find the allele. Write down the genotypes of the parents. List all the possible gamete types. Connect. Determine cleavage.

In problems of the second type, the opposite is true. Splitting in offspring is known here. It is required to determine the genotypes of the parents. Find, as in the first type of problems, which of the alleles is dominant and which is recessive. Identify possible gamete types. Use them to determine the genotypes of the parents.

To solve the problem correctly, read it carefully and analyze the condition. To determine the type of problem, find out how many feature pairs are considered in the problem. Also notice how many pairs of genes control the development of traits. It is important to find out if homozygous or crossed, what is the type of crossing. Determine whether genes are independently or linked, how many genotypes are formed in the offspring, and whether inheritance is sex-related.

Start solving the problem. Make a short note of the condition. Record the genotype or phenotype of the individuals involved in the crossing. Identify and mark the types of gametes produced. Record the genotypes or phenotypes of the offspring resulting from the crossing. Analyze the results, write them down numerically. Write the answer.