The assignment is for top level difficulties. For the correct answer you will receive 3 points.

Approximately up to 10-20 minutes.

To complete task 28 in biology, you need to know:

• how (compose crossbreeding schemes), ecology, evolution;

Tasks for training

Task #1

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

Task #2

In Drosophila, the black color of the body dominates over the gray, normal wings - over the 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 #3

A person has 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 gene for color blindness d is linked to the X chromosome. A woman with blood group 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, the mother was healthy. The man's relatives never had this disease. Determine the genotypes of the parents. Specify the possible genotypes and phenotypes (blood group number) of children. Make a scheme for solving the problem. Determine the probability of the birth of children with color blindness and children with II blood group.

Task #4

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

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

The average general education

Line UMK VV Pasechnik. Biology (10-11) (base)

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

Biology

USE in biology-2018: task 27, basic level

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

Task 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 options for tasks on gene mutations, but it is also appropriate to repeat chromosomal and genomic mutations.

In general, task number 27 this year is represented by 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 polypeptide fragment. Regardless of the wording, it is extremely important to remind students how to write DNA nucleotide sequences correctly: without spaces, hyphens and commas, with 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 a mutation if one amino acid is replaced by another in a protein? 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 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 should be studied in order to successfully solve the problems of the 27th line?

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


• Microsporogenesis and macrosporogenesis in gymnosperms and angiosperms.

The attention of students and teachers is offered a new tutorial which will help to successfully prepare for a single state exam in biology. The collection contains questions selected by sections and topics tested at the exam, and includes tasks of different types and levels of complexity. Answers to all questions are given at the end of the manual. The proposed thematic tasks will help the teacher organize preparation for the unified state exam, and students will independently test their knowledge and readiness for the final exam. The book is addressed to students, teachers and methodologists.

How to prepare?

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


2. Turn on visual memory. It is useful to memorize illustrations of the basic schemes of the evolution of the life cycle of various plants - for example, the cycle of alternation of generations in algae, ferns, bryophytes. Unexpectedly, but questions related to the life cycle of pine, for some reason, 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 must be understood that the bump itself is diploid: for the teacher this is obvious, but for the student it is not always.


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, an alternation of a haploid gametophyte and a diploid sporophyte is observed, with the latter predominating (this way we will get rid of possible nit-picking). A nuance in the topic of the life cycle of ferns: explaining from 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 spore mother cells. Both answers are satisfactory.

We analyze examples of tasks

Example 1 A fragment of a DNA chain has the following sequence: TTTGCGATGCCCCA. Determine the sequence of amino acids 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 has been 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 your answer using the genetic code table.

Solution. This task is easily decomposed into elements that will make up the correct answer. It is best to act according to a proven algorithm:

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


2. 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 (here, the skill of solving such problems is needed).

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 meiosis, in meiosis anaphase I and meiosis anaphase II. Explain what processes take place 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 you to determine the chromosome set and the number of DNA molecules, besides, they show numbers - do not limit yourself to the formula: be sure to indicate the numbers.

The following steps are required for the solution:

1. indicate the initial number of DNA molecules. In this case, it is 56 - since they double, 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 is 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.

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

Example 3 What chromosome set is typical for 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 the interphase before the beginning of division and after the 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-respectively 60 and 30. It is important to note in the answer that before the start of division, the DNA molecules are doubled, and 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 typical for the cells of the outgrowth and gametes of the fern? 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 is easily decomposed into three elements:

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


2. be sure to indicate that the outgrowth develops from a haploid spore by mitosis, and gametes - on a haploid outgrowth, 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 somatic cells. Determine the chromosome set and the number of DNA molecules in cells before meiosis, in meiosis anaphase I and in meiosis prophase II. Explain the answer.

Solution. As you can see, in such tasks, the number of response criteria is clearly 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 in the answer to give explanations after each numerical answer. For instance:

1. we give the formula: before meiosis, the set of chromosomes and DNA is 2n4c; at the end of the interphase, DNA doubling occurred, the chromosomes became two-chromatid; 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. meiotic prophase II is entered by haploid cells having a set of two-chromatid chromosomes with a set of n2c. Therefore, at this stage we have 24 chromosomes and 48 DNA molecules.

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

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

Description of the presentation on individual slides:

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Municipal budgetary educational institution"Karpovskaya average comprehensive school» Urensky municipal district Nizhny Novgorod Region "Analysis of the 28th task of the Unified State Examination 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. The task of genetics. Genealogical method The genealogical method consists in the analysis of pedigrees and allows you to determine the type of inheritance (dominant recessive, autosomal or sex-linked) of a trait, as well as its monogenicity or polygenicity. The person in respect of whom a pedigree is made is called a proband, and his brothers and sisters of the proband are called sibs.

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Task 28. The task of 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. Patients occur in every generation. 2. Get sick equally both men and women. 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 in healthy parents is 0%. Autosomal recessive type of inheritance. 1. Patients are not found in every generation. 2. Both men and women get sick 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 manifested in closely related marriages.

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

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Types of trait inheritance Hollandic type of inheritance (Y-linked inheritance). 1. Patients occur in every generation. 2. Only men get sick. 3. If a father is sick, then all his sons are sick. 4. The probability of the birth of a sick boy from a sick father is 100%.

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Task 28. The task of genetics. Genealogical method Stages of problem solving Determine the type of trait inheritance - 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 sign, if the parents do not show this sign? Are there cases of children being born without the trait being studied, if both parents have it? What part of the offspring carries a trait in families if one of the parents is its owner? 2. Determine if the trait is inherited sex-linked. how often does the symptom occur in both sexes (if it is rare, then which sex has it more often)? Which sexes inherit the trait from the father and mother who carry 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. The task of genetics. Genealogical method Task 1. According to the pedigree shown in the figure, establish the nature of the inheritance of the trait highlighted in black (dominant or recessive, sex-linked or not), the genotypes of children in the first and second generation.

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Task 28. The task of 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) 2. Determine whether the trait is inherited gender, because the trait is transmitted equally to both sons and daughters). 3. We determine the genotypes of the parents: (female aa (without a sign of homozygous), man Aa (with a sign) is a 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 trait is dominant, since it is always transmitted to offspring, not sex-linked, since it is transmitted equally to both daughters and sons. Genotypes of parents: female: aa, male Aa (with trait). 2) Genotypes of children in F1 women - Aa (with a trait) and aa, men - Aa (with a trait). 3) Genotypes of descendants F2 male - Aa (with trait).

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

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Task 28. The task of genetics. Genealogical method Solution algorithm 1. Determine the type of inheritance: the trait occurs in all generations or not (the trait is recessive, because it is not present in all generations) 2. Determine the genotypes of the parents: (male Aa (without a trait), female aa (with a trait) 3. We solve the problem with genotypes: P: aa (f with a sign) x Aa (m. without a sign) G: a A a F1: Aa (m. without a sign), Aa (f. without a sign) 4. Write down the answer : 1) The trait is recessive; 2) genotypes of parents: mother - aa, father - AA or Aa; 3) genotypes of children: son and daughter of heterozygotes - Aa (allowed: other genetic symbolism that does 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. The task of genetics. Genealogical method Task 3. According to the pedigree shown in the figure, determine and explain the nature of the inheritance of the trait (dominant or recessive, sex-linked or not), highlighted in black. Determine the genotypes of the offspring indicated in the diagram by the numbers 3, 4, 8, 11 and explain the formation of their genotypes.

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Task 28. The task of genetics. Genealogical method Solution algorithm 1. Determine the type of inheritance: the trait occurs in all generations or not (the trait is recessive, because it is not present in all generations) 2. Determine whether the trait is inherited sex-linked: more common in boys or girls (linked with the X chromosome, because there is a slip through the generation). 3. We determine the genotypes of people indicated on the diagram by the numbers 3, 4, 8, 11: 4. Write down the answer. 3 - female carrier - HAHA 4 - male without symptom - XAY 8 - male with symptom - XAY 11 - female carrier - HAHA

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Task 28. The task of genetics. Genealogical method Task 4. Determine the type of inheritance, the genotype of the proband in the following pedigree. Determining the type of inheritance of the trait: The studied trait 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 located on the Y chromosome. In women, this trait is absent, since the pedigree shows that the trait is not transmitted through the female line. Therefore, the type of inheritance of the trait: linked to the Y-chromosome, or hollandric inheritance of the trait. 1. the trait occurs frequently, in every generation; 2. the sign occurs only in men; 3. the trait 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; YA - normal development of the organism (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 hollandic inheritance. Proband genotype: XYa.

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Task 28. The task of genetics. Codominance. Interaction of genes. Task 1. The color gene of cats is linked to the X chromosome. Black coloration is determined by the XA gene, red - by the XB gene. Heterozygotes are tortoiseshell. From a tortoiseshell cat and a red cat, five red kittens were born. Determine the genotypes of parents and offspring, the nature of the inheritance of traits. Solution algorithm: Let's write down the condition of the problem: XA - black; ХВ - red, then ХХВ - tortoiseshell 2. Let's write down the genotypes of the parents: Р: cat ХАХB x cat ХВУ turtles. red G: XA XB XB Y.F1: red - XBY or XBXB sex-linked inheritance

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Task 28. The task of genetics. Codominance. Interaction of genes. Task 2. Cat coat color genes are located on the X chromosome. Black color is determined by the gene XB red - Xb, heterozygotes are tortoiseshell. From a black cat and a red cat, one tortoiseshell and one black kitten were born. Determine the genotypes of parents and offspring, the possible sex of the kittens. Solution algorithm: Let's write down the crossbreeding scheme, the genotype of a black cat is XB XB, the genotype of a red cat is Xb Y, the genotypes of kittens are: tortoiseshell - XB Xb, black - XB Y, the sex of kittens: tortoiseshell - female, black - male.

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Task 28. The task of genetics. Codominance. Interaction of genes. Task 3. A person has four phenotypes according to blood groups: 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 (heterozygous) and III (homozygous) blood groups. Determine the genotypes of the blood groups of the parents. Specify the possible genotypes and phenotypes (number) of the blood group of children. Make a scheme for solving the problem. Determine the probability of inheritance in children of the II blood group. Solution algorithm: 1) parents have blood types: group II - IAi0 (gametes IA, i0), group III - IVIB (gametes IB); 2) possible phenotypes and genotypes of children's blood groups: group IV (IAIB) and group III (IBi0); 3) the probability of inheritance of blood group II - 0%.

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Task 28. The task of genetics. Mono- and dihybrid crossing Problem 1. When crossing a corn plant with smooth colored seeds and a plant with wrinkled uncolored seeds, all hybrids of the first generation had smooth colored seeds. From analyzing crosses of F1 hybrids obtained: 3800 plants with smooth colored seeds; 150 - with wrinkled dyed; 4010 - with wrinkled unpainted; 149 - with 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 four phenotypic groups in test crosses.

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Task 28. The task of genetics. Mono- and dihybrid cross Solution algorithm: 1) First cross: P AABB × aabb G AB × ab F1 AaBb 2) Test cross: P AaBb × aabb G AB, Ab, aB, ab × ab AaBb - 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. The task of genetics. Mono- and dihybrid crossing Task 2. When crossing white guinea pigs with smooth hair with black guinea pigs with shaggy hair, offspring were obtained: 50% black shaggy and 50% black smooth. When crossing the same smooth-coated white gilts with other black shaggy-haired gilts, 50% of the offspring were black shaggy and 50% were white shaggy. Make a diagram of each cross. Determine the genotypes of parents and offspring. What is the name of this crossing and why is it carried out?

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Task 28. The task of genetics. Mono- and dihybrid crossing Task 3. In peas, the pink color of the corolla dominates over the white one, and the tall stem dominates 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 dwarf stems. Make a scheme for solving the problem. Determine the genotypes of the original plants and descendants. Explain the nature of the inheritance of traits and the formation of four phenotypic groups.

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

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Task 28. The task of genetics. Linkage of genes Task 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 probable genotypes of all children, parents, as well as the possible genotypes of the grandfathers of these children. Solution algorithm 1) Parents with normal vision: father ♂ХDY, mother ♀ХDХd. 2) Gametes ♂ XD, Y; ♀ Xd, XD. 3) Possible genotypes of children - daughters X DXd or XDX D; sons: color-blind XdY and son with normal vision X DY. 4) Grandfathers or both are color blind - XdY, or one XDY and the other XdY.

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Task 28. The task of genetics. Linkage of genes Problem 2. A woman with a recessive gene for hemophilia married a healthy man. Determine the genotypes of the parents, and the expected offspring - the ratio of genotypes and phenotypes. Solution algorithm 1) Genotypes of parents CN Xh and CNV; 2) offspring genotypes - XH 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, sick with hemophilia. Phenotype ratio 2 (healthy girls): 1 (healthy boy) : 1 (hemophilic boy)

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Task 28. The task of genetics. Gene linkage Task 3. In humans, the inheritance of albinism is not sex-linked (A - the presence of melanin in skin cells, a - 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 the marriage of a dihomozygous normal woman and albino man with hemophilia for both alleles. Make a scheme for solving the problem. Solution algorithm 1) parental genotypes: ♀AAXHXH (AXH gametes); ♂aaXhY (gametes aXh, aY); 2) genotypes and sex of children: ♀AaXHXh; ♂AaXHY; 3) phenotypes of children: outwardly normal for both alleles, a girl, but a carrier of albinism and hemophilia genes; outwardly normal for both alleles, a boy, but a carrier of the albinism gene.

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

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

The questions of task 28 of the USE in biology may also affect the basics of biology. Be sure to repeat before the exams - what is the absence of artificial and natural ecosystems, water and terrestrial, meadow and field, how the rule of the ecological pyramid sounds and where it applies, what is biogeocenosis and agrocenosis. Some questions are logical in nature, you need to not only rely on the theory from a school textbook, but also 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. Which one? May beetle, bird cherry, mushrooms, wild rose, hazel, mice are offered as answers. In this case, the student must remember that competition always goes for the same resources, in this case (with a tiered arrangement of plants) for light, so only trees and shrubs need to be selected from the list - bird cherry, wild rose and hazel.

Based on the pedigree shown in the figure, determine and explain the nature of the inheritance of the trait (dominant or recessive, sex-linked or not), highlighted in black. Determine the genotypes of the offspring indicated in the diagram by the 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 "slip" through the 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 Xa, the other XA.

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

3 - female carrier - X A X a

4 - a man without a sign - X A Y

8 - a man with a 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 a number) HAHA, because she can be a carrier?

Natalya Evgenievna Bashtannik

Maybe. This is a "guess" based offspring. Because it is not important for us to solve, you can write both options on the diagram, or you can do it like this: Х А Х -

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, heterozygotes Aa, men 4 and 5 are also carriers of Aa, children 7 and 8 in the second generation with a 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, one without, possibly a carrier.

Natalya Evgenievna Bashtannik

maybe, but more likely that there is a chain, less "?", and based on the rules for solving these problems.

A SON with a trait is born in a mother and father phenotypically without a 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 less assumptions:

In fact, all we can assert from the task data is a list of what we can rule out. We can rule out dominant linkage to X, we can rule out linkage to Y, we can rule out AA x aa in the Persian cross itself, we can rule out that the trait is provided by the dominant allele.

We can't rule out recessive linkage to X, and we can't rule out autosomal recessive inheritance—there's not enough data in the problem for that, and not enough offspring and crosses.

To ignore the small number of crosses and offspring is to assume that the law of large numbers must also apply to small numbers. What complete nonsense. Should not. On the contrary: the statistical fact is that the smaller the sample, the greater the expected deviation from the "correct splitting".

Natalya Evgenievna Bashtannik

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