For this task, you can get 3 points on the exam in 2020

The topic of the 28th Unified State Exam in biology was "Superorganismic systems and the evolution of the world." Many schoolchildren note the difficulty of this test due to the large volume of educational material covered by it, as well as due to the construction of the ticket. In task number 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 signs out of the six listed are characteristic of animal cells"), but in most cases the student must make his own decision about the number of answers he chooses as correct.

The questions of the 28 exam in biology can also affect the basics of biology. Be sure to repeat before exams - what is the absence of artificial and natural ecosystems, water and terrestrial, meadow and field, how does the rule of an ecological pyramid sound 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 a birch and another living organism. How? " May beetle, bird cherry, mushrooms, wild rose, hazel, mice are offered as answers. In this case, the student must remember that the competition always goes for the same resources, in this case (with a tiered arrangement of plants) - for light, therefore only trees and shrubs need to be selected from the list - bird cherry, wild rose and hazel.

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 of the same 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 green and wrinkled.

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 a dominant gene, others a recessive one. 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, and the parental forms are heterozygous for this trait (because 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 a 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. They 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

V.V. Pasechnik's UMK line. Biology (10-11) (basic)

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

Biology

Unified State Exam in Biology-2018: task 27, basic level

Assignment 27: What's New?

Part of the tasks of this line has changed: now it is more often required to predict the consequences of a mutation of a gene section. First of all, there will be variants of tasks for gene mutations, but chromosomal and genomic mutations are also appropriate to repeat.

In general, the task number 27 this year is presented with very different options. Some of the tasks are related to protein synthesis. It is important to understand here: the algorithm for solving the 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 can be proposed and simply to synthesize a fragment of the polypeptide. Regardless of the wording, it is extremely important to remind students how to write DNA nucleotide sequences correctly: no spaces, hyphens or commas, a continuous sequence of characters.

To correctly solve problems, 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 the DNA fragment in accordance with the mutation.

If the problem contains the formulation “explain using your knowledge of 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, plant development cycles: algae, mosses, ferns, gymnosperms, angiosperms.


• Microsporogenesis and macrosporogenesis in gymnosperms and angiosperms.

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 to pass the final exam. The book is addressed to students, teachers and methodologists.

How to prepare?

1. Show students the schemes and algorithms: as in plants spores and gametes are formed, as in animals - gametes and somatic cells. It is useful to ask students to model their own schemes of mitosis and meiosis: this allows you to understand why haploid cells formed during meiosis later become diploid.


2. Turn on visual memory. It is useful to memorize illustrations of the basic evolutionary patterns of the life cycle of various plants - for example, the cycle of alternation of generations in algae, ferns, bryophytes. Unexpectedly, for some reason, issues related to the life cycle of a 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 bump itself is diploid: this is obvious for the teacher, but not always for the student.


3. Pay attention to the nuances of the wording. When describing some issues, clarifications need to be made: 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 nagging). A nuance in the topic of the life cycle of ferns: explaining from what and how disputes are formed, you can answer in different ways. One option is from sporagon cells, and the other, more convenient, from spore mother cells. Both answers are satisfactory.

We analyze examples of tasks

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

Solution. This problem can be easily decomposed into elements that will make up the correct answer. It is best to follow the proven algorithm:

1. determine the amino acid sequence in the fragment;


2. we write what happens when one amino acid is replaced;


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

Example 2. The chromosome set of wheat somatic cells is 28. Determine the chromosome set and the number of DNA molecules in one of the ovule cells before the onset of meiosis, in meiosis anaphase I and meiosis anaphase II. Explain what processes occur during these periods and how they affect the change 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 task asks to determine the chromosome set and the number of DNA molecules, and besides, they show 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 are doubled, and the number of chromosomes does not change;


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


3. describe the anaphase of meiosis II: the number of DNA molecules - 28, chromosomes - 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.

With this wording, the answer is likely to yield the desired high score.

Example 3. What chromosome set is typical for the cells of the pollen grain and sperm cells of the pine? 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 can be 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 the cells of the pollen grain (generative cells), also by mitosis.

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

Solution. The problem is solved according to the previously described algorithm. In the 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 the start of division, DNA molecules are doubled, and the number of chromosomes does not change; we are dealing with reduction division, so the number of DNA is halved.

Example 5. What chromosomal set is typical for the cells of the germ and fern 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 be easily decomposed into three elements:

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


2. we must point out that the germ develops from a haploid spore by mitosis, and gametes - on a haploid germ, by mitosis;


3. since the exact number of chromosomes is not indicated, you can limit yourself to the formula and write just 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 the anaphase of meiosis I and in the prophase of meiosis II. Explain the answer.

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

1. we give the formula: before the onset of meiosis, the set of chromosomes and DNA is equal to 2n4c; at the end of the interphase, DNA was doubled, chromosomes became dichromatid; 48 chromosomes and 96 DNA molecules;


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


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

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 reference book 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 assignments correspond to the format of the Unified State 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.

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

Description of the presentation for individual slides:

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Municipal budgetary educational institution "Karpovskaya secondary school" of the Uren municipal district of the Nizhny Novgorod region "Analysis of 28 tasks of the exam in biology, part C" Prepared by: teacher of biology and chemistry MBOU "Karpovskaya secondary school" Chirkova Olga Alexandrovna 2017

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Task 28. Problem in genetics. Genealogical method The genealogical method consists in the analysis of pedigrees and allows to determine the type of inheritance (dominant recessive, autosomal or sex-linked) of the trait, as well as its monogenicity or polygenicity. The person in relation to whom the pedigree is made up is called the proband, and his brothers and sisters are called the proband siblings.

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Task 28. Problem in genetics. Genealogical method Symbols used in the compilation of pedigrees

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Types of inheritance of traits Autosomal dominant type of inheritance. 1. Sick people are found in every generation. 2. Both men and women are ill equally. 3. A sick child is born to sick parents with a probability of 100% if they are homozygous, 75% if they are heterozygous. 4. The probability of having a sick child among healthy parents is 0%. Autosomal recessive inheritance. 1. Patients are not found in every generation. 2. Both men and women are ill equally. 3. The probability of having a sick child in healthy parents is 25%, if they are heterozygous; 0% if both, or one of them, are homozygous for the dominant gene. 4. Often manifests itself in closely related marriages.

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Types of inheritance of traits X-linked (sex) dominant type of inheritance 1. Patients are found in every generation. 2. Mostly women are ill. 3. If a father is sick, then all his daughters are sick. 4. A sick child is born to sick parents with a 100% probability if the mother is homozygous; 75% if the mother is heterozygous. 5. The probability of having a sick child from healthy parents is 0%. X-linked (sex) recessive inheritance. 1. Patients are not found in every generation. 2. Mostly men are ill. 3. The probability of having a sick boy in healthy parents is 25%, a sick girl is 0%.

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Types of inheritance of traits Hollandric type of inheritance (Y-linked inheritance). 1. Sick people are found in every generation. 2. Only men are ill. 3. If a father is sick, then all his sons are sick. 4. The probability of having a sick boy from a sick father is 100%.

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Task 28. Problem in genetics. Genealogical method Stages of problem solving Determine the type of inheritance of the trait - dominant or recessive. Answer the questions: Does the trait occur in all generations or not? How often does the trait occur in members of the pedigree? Are there cases of the birth of children with a trait, if the parents do not show this trait? Are there cases of the birth of children without the studied trait, if both parents have it? What part of the offspring bears the trait in families, if one of the parents is its owner? 2. Determine whether the trait is inherited is sex-linked. How common is the symptom in persons of both sexes (if it is rare, then the persons of which sex carry it more often)? persons of which gender inherit the trait from the father and mother carrying the trait? 3. Find out the formula for splitting offspring in one generation. And based on the analysis, determine the genotypes of all members of the pedigree.

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Task 28. Problem in genetics. Genealogical method Task 1. According to the pedigree shown in the figure, establish the nature of inheritance of the trait highlighted in black (dominant or recessive, linked or not linked to the sex), genotypes of children in the first and second generation.

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Task 28. Problem in genetics. Genealogical method Decision algorithm 1. Determine the type of inheritance: the trait occurs in all generations or not (the trait is dominant, because it is always transmitted to offspring) gender, since the trait is transmitted equally to sons and daughters). 3. Determine the genotypes of the parents: (woman aa (without a sign of homozygote), man Aa (with a sign) - heterozygote. 4. Solve the problem with genotypes: P: aa (g) x Aa (m with a sign) G: a A a F1: Aa (m. With a sign), Aa (f. With a sign), aa (f. Without a sign) P: Aa (f. With a sign) x aa (m. Without a sign) F2: Aa (m. With a sign) ) 5. We write down the answer: 1) The dominant trait, since it is always transmitted to the offspring, is not sex-linked, since it is transmitted equally to both daughters and sons. Genotypes of parents: woman: aa, man Aa (with a trait). 2) Genotypes of children in F1 women - Aa (with a trait) and aa, men - Aa (with a trait). 3) Genotypes of offspring F2 male - Aa (with a trait).

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Task 28. Problem in genetics. Genealogical method Task 2. Using the pedigree shown in the figure, establish the nature of the manifestation of the trait (dominant, recessive), indicated in black. Determine the genotype of the parents and children in the first generation.

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Task 28. Problem in genetics. Genealogical method Decision algorithm 1. Determine the type of inheritance: the trait occurs in all generations or not (the trait is recessive, because we are not present in all generations) 2. Determine the genotypes of the parents: (male Aa (without a trait), woman aa (with a trait) 3. Solve the problem with genotypes: P: aa (w with a sign) x Aa (m without a sign) G: a A a F1: Aa (m without a sign), Aa (w. Without a sign) 4. Write down the answer : 1) The sign is recessive; 2) genotypes of parents: mother - aa, father - AA or Aa; 3) the genotypes of children: the son and daughter of the heterozygote - Aa (it is allowed: other genetic symbols that do not distort the meaning of solving the problem, indicating only one of the variants of the father's genotype).

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Task 28. Problem in genetics. Genealogical method Task 3. Using the pedigree shown in the figure, determine and explain the nature of inheritance of the trait (dominant or recessive, linked or not with sex), highlighted in black. Determine the genotypes of the offspring indicated on the diagram by the numbers 3, 4, 8, 11 and explain the formation of their genotypes.

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Task 28. Problem in genetics. Genealogical method Solution algorithm 1. Determine the type of inheritance: the trait occurs in all generations or not (the trait is recessive, because we are not present in all generations) with the X - chromosome, because there is a breakthrough through the generation). 3. Determine the genotypes of people indicated on the diagram by the numbers 3, 4, 8, 11: 4. Write down the answer. 3 - a female carrier - HAH 4 - a man without a sign - HAY 8 - a man with a sign - HAY 11 - a woman carrier - HAH

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Task 28. Problem in genetics. Genealogical method Task 4. Determine the type of inheritance, the genotype of the proband in the following pedigree Determine the type of inheritance of the trait: The trait under study is found only in males in each generation and is transmitted from father to son (if the father is sick, then all sons also suffer from this disease), then we can think that the gene under study is on the Y-chromosome. In women, this trait is absent, since it is clear from the pedigree that the trait is not transmitted along the female line. Therefore, the type of inheritance of the trait: linked to the Y-chromosome, or holandric inheritance of the trait. 1. the sign occurs frequently, in every generation; 2. the symptom is found only in men; 3.the sign is transmitted through the male line: from father to son, etc. Possible genotypes of all members of the pedigree: Ya - the presence of this anomaly; YА - normal development of the body (absence of this anomaly). All men suffering from this anomaly have the genotype: XYa; All men who do not have this anomaly have the genotype: XYA. Answer: Linked to the Y chromosome, or Dutch inheritance. Proband genotype: XYa.

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Task 28. Problem in genetics. Codominance. Interaction of genes. Problem 1. The gene for the color of cats is linked to the X chromosome. Black color is determined by the XA gene, red color - by the XB gene. Heterozygotes have a tortoiseshell coloration. Five ginger kittens were born from a tortoiseshell cat and a ginger cat. Determine the genotypes of parents and offspring, the nature of inheritance of traits. Algorithm for solving: Let's write down the condition of the problem: ХА - black; ХВ - red, then ХАХВ - tortoiseshell 2. Let's write down the genotypes of the parents: P: cat ХАХB х cat ХБУ turtles. redhead G: XA XB XB U. F1: red - XBY or XBXB inheritance linked to sex

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Task 28. Problem in genetics. Codominance. Interaction of genes. Problem 2. Genes for the color of cats' fur are located on the X chromosome. Black color is determined by the gene XB red - Xb, heterozygotes have a tortoiseshell color. One tortie and one black kitten were born from a black cat and a ginger cat. Determine the genotypes of the parents and offspring, the possible sex of the kittens. Algorithm for the solution: Let us write down the crossing scheme: the genotype of the black cat is XB XB, the genotype of the red cat is Xb Y, the genotypes of the kittens: tortoiseshell - XB Xb, Black - XB Y, sex of kittens: tortie - female, black - male.

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Task 28. Problem in genetics. Codominance. Interaction of genes. Problem 3. A person has four phenotypes by blood group: I (0), II (A), III (B), IV (AB). The gene that determines the blood group has three alleles: IA, IB, i0, and the i0 allele is recessive with respect to the IA and IB alleles. Parents have II (heterozygote) and III (homozygote) blood groups. Determine the genotypes of the parents' blood groups. Indicate the possible genotypes and phenotypes (number) of the blood group of children. Make a scheme for solving the problem. Determine the likelihood of inheritance in children of the II blood group. Solution algorithm: 1) parents have blood groups: II group - IAi0 (gametes IA, i0), III group - IVIB (gametes IV); 2) possible phenotypes and genotypes of blood groups of children: group IV (IAIB) and group III (IBi0); 3) the probability of inheriting blood group II is 0%.

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Task 28. Problem in genetics. Mono- and dihybrid crossing Problem 1. When crossing a maize plant with smooth colored seeds and plants with wrinkled uncolored seeds, all hybrids of the first generation had smooth colored seeds. Analyzing crosses of F1 hybrids yielded: 3800 plants with smooth colored seeds; 150 - wrinkled colored; 4010 - wrinkled, unpainted; 149 - smooth, unpainted. Determine the genotypes of the parents and offspring obtained as a result of the first and analyzing crosses. Make a scheme for solving the problem. Explain the formation of the four phenotypic groups in the analyzing cross.

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Task 28. Problem in genetics. Mono- and dihybrid crossing Solution algorithm: 1) First crossing: P ААBB × ааbb G AB × ab F1 AaBb 2) Analyzing crossing: P AaBb × ааbb G AB, Ab, aВ, ab × ab АаBb - smooth colored seeds (3800) ; Aabb - smooth uncolored seeds (149); aaBb - wrinkled colored seeds (150); aabb - wrinkled uncolored seeds (4010); 3) the presence in the offspring of two groups of individuals with dominant and recessive traits in approximately equal proportions (3800 and 4010) is explained by the law of linked inheritance of traits. Two other phenotypic groups (149 and 150) are formed as a result of crossing over between allelic genes.

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Task 28. Problem in genetics. Mono- and dihybrid crossing Problem 2. When crossing white guinea pigs with smooth hair with black pigs with shaggy hair, the offspring were obtained: 50% black shaggy and 50% black smooth. When crossing the same white pigs with smooth hair with other black pigs with shaggy hair, 50% of the offspring were black shaggy and 50% - white shaggy. Make a diagram of each cross. Determine the genotypes of the parents and offspring. What is the name of this crossing and why is it carried out?

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Task 28. Problem in genetics. Mono- and dihybrid crossing Problem 3. In seed peas, the pink color of the corolla dominates over the white one, and the tall stem over the dwarf one. When a plant with a high stem and pink flowers was crossed with a plant with pink flowers and a dwarf stem, 63 plants with a high stem and pink flowers were obtained, 58 - with pink flowers and a dwarf stem, 18 - with white flowers and a high stem, 20 - with white flowers and a dwarf stem. Make a scheme for solving the problem. Determine the genotypes of the original plants and offspring. Explain the nature of the inheritance of traits and the formation of four phenotypic groups.

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Task 28. Problem in genetics. Mono- and dihybrid crossing Solution algorithm: 1) P AaBb x Aabb pink flowers pink flowers high stem high stem G AB, Ab, aB, ab Ab, ab 2) F1 AaBb, AABb - 63 pink flowers, high stem Aabb, AAbb - 58 pink flowers, dwarf stem aaBb - 18 white flowers, high stem aabb - 20 white flowers, dwarf stem. 3) The genes of two traits are not linked with complete dominance, therefore the inheritance of traits is independent.

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Task 28. Problem in genetics. Coupling of genes Problem 1. A married couple, in which both spouses had normal vision, were born: 2 boys and 2 girls with normal vision and a color-blind son. Determine the likely genotypes of all children, the parents, and the possible genotypes of the grandfathers of these children. Solution algorithm 1) Parents with normal vision: father ♂XDY, mother ♀XDXd. 2) Gametes ♂ XD, Y; ♀ Xd, XD. 3) Possible genotypes of children - daughters X DXd or XX D; sons: color blind XdY and son with normal vision X DY. 4) Grandfathers or both are color blind - XdY, or one is XDY, and the other is XdY.

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Task 28. Problem in genetics. Genes clutch Problem 2. A woman with a recessive hemophilia gene married a healthy man. Determine the genotypes of the parents, and in the expected offspring - the ratio of genotypes and phenotypes. Algorithm for solving 1) Genotypes of parents XN Xh and XHY; 2) genotypes of offspring - XN Xh, XN XH, XN Y, XhY; The ratio of genotypes is 1: 1: 1: 1 3) daughters are a carrier of the hemophilia gene, healthy, and sons are healthy, have hemophilia. Ratio of phenotypes 2 (girls are healthy): 1 (boy is healthy): 1 (boy is hemophilic)

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Task 28. Problem in genetics. Genetic linkage Problem 3. In humans, the inheritance of albinism is not sex-linked (A - the presence of melanin in skin cells, and - the absence of melanin in skin cells - albinism), and hemophilia is sex-linked (XH - normal blood clotting, Xh - hemophilia) ... Determine the genotypes of the parents, as well as the possible genotypes, sex and phenotypes of children from a marriage of dihomozygous normal for both alleles of an albino woman and a man with hemophilia. Make a scheme for solving the problem. Solution algorithm 1) parental genotypes: ♀AAXHXH (AXH gametes); ♂aaXhY (gametes aXh, aY); 2) genotypes and gender of children: ♀AaXHXh; ♂AaXHY; 3) phenotypes of children: a girl outwardly normal in both alleles, but a carrier of genes for albinism and hemophilia; a boy outwardly normal in both alleles, but a carrier of the albinism gene.