Biology problems 28 with solutions. Assignments on genetics for the Unified State Exam in biology. Task C6. Problems to solve independently
The assignment relates to highest level difficulties. For the correct answer you will receive 3 points.
It takes approximately up to 10-20 minutes.
To complete task 28 in biology you need to know:
- how to (make crossbreeding schemes), ecology, evolution;
Tasks for training
Task No. 1
The hamster color gene is linked to the X chromosome. The X A genome is determined by the brown color, the X B genome is black. Heterozygotes have a tortoiseshell coloration. Five black hamsters were born from a female tortoiseshell and a black male. Determine the genotypes of parents and offspring, as well as the nature of inheritance of traits.
Task No. 2
In fruit flies, black body color dominates over gray, and normal wings dominate over curved ones. Two black flies with normal wings are crossed. The offspring of F 1 are phenotypically uniform - with a black body and normal wings. What are the possible genotypes of the crossed individuals and offspring?
Task No. 3
Humans have four phenotypes according to blood groups: I(0), II(A), III(B), IV(AB). The gene that determines the blood group has three alleles: I A, I B, i 0, and the i 0 allele is recessive with respect to the IA and IB alleles. The color blindness gene d is linked to the X chromosome. A woman with blood type II (heterozygote) and a man with blood group III (homozygote) entered into marriage. It is known that the woman’s father suffered from color blindness, her mother was healthy. The man's relatives never had this disease. Determine the genotypes of the parents. Indicate possible genotypes and phenotypes (blood group number) of children. Make a diagram for solving the problem. Determine the probability of having color-blind children and children with blood group II.
Task No. 4
In corn, the genes for brown color and smooth seed shape are dominant over the genes for white color and wrinkled shape.
When plants with brown smooth seeds were crossed with plants with white seeds and wrinkled seeds, 4006 brown smooth seeds and 3990 white wrinkled seeds were obtained, as well as 289 white smooth and 316 brown wrinkled corn seeds. Make a diagram for solving the problem. Determine the genotypes of the parent corn plants and their offspring. Justify the appearance of two groups of individuals with characteristics different from their parents.
Using the pedigree shown in the figure, determine and explain the nature of inheritance of the trait (dominant or recessive, sex-linked or not) highlighted in black. Determine the genotypes of the descendants indicated in the diagram by numbers 3, 4, 8, 11 and explain the formation of their genotypes.
Explanation.
The trait highlighted in black is recessive, linked to the X chromosome: X a,
because there is a “leap” through a generation. A man with a sign (8) has a daughter without a sign (11), and grandchildren - one with a sign (12), the second without (13), that is, they receive a Y chromosome from their father (10), and from their mother (11) one X a, the other X A.
Genotypes of people indicated in the diagram by numbers 3, 4, 8, 11:
3 - female carrier - X A X a
4 - man without sign - X A Y
8 - a man with the sign - X and Y
11 - female carrier - X A X a
Source: Unified State Examination in Biology 05/30/2013. Main wave. Far East. Option 4.
Elena Ivanova 11.04.2016 12:36
Please explain why the genotype of the first woman (without number) is HAHA, because she could also be a carrier?
Natalia Evgenievna Bashtannik
Maybe. This is a "guess" based on offspring. Because it is not important for us to solve, we can write both options on the diagram, or even like this: X A X -
Nikita Kaminsky 11.06.2016 23:28
Why can’t there just be a recessive gene that is not sex-linked?
Then the parents in the first generation are homozygous (father aa, mother AA), children 1, 2, 3 are heterozygous Aa, men 4 and 5 are also carriers of Aa, children 7 and 8 in the second generation have the trait, and 6 is a carrier. In the third generation, Father and Mother are again homozygous, daughter 11 and her husband 10 are heterozygous, and they have two sons, one with the trait, the other without, possibly a carrier.
Natalia Evgenievna Bashtannik
maybe, but there is a greater probability that there is clutch, less “?”, and based on the rules for solving these problems.
A mother and father who are phenotypically without the trait give birth to a SON with the trait; it can be assumed that the trait is linked to the X chromosome.
Tobias Rosen 09.05.2017 18:26
The solution is not entirely correct.
This diagram contains an alternative solution - containing fewer assumptions:
In fact, all we can say based on the problem data is a list of what we can exclude. We can exclude dominant linkage to X, we can exclude linkage to Y, we can exclude AA x aa in the cross itself, we can exclude that the trait is provided by a dominant allele.
We cannot exclude recessive linkage with X and we cannot exclude autosomal recessive inheritance - there is not enough data for this and an insufficient number of descendants and crosses.
To ignore the small number of crossings and descendants is to assume that the law of large numbers must also apply to small numbers. Which is complete nonsense. Should not. On the contrary: the statistical fact is that the smaller the sample, the greater the expected deviation from the “correct split.”
Natalia Evgenievna Bashtannik
If a problem can be solved in two ways, then it is better to write both. If the criteria include a decision that the trait is linked to the X chromosome: X a, then they may not give a full point.
Description of the presentation by individual slides:
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Municipal budget educational institution"Karpovskaya average comprehensive school» Urensky municipal district Nizhny Novgorod region "Parsing 28 Unified State Exam assignments in biology, part C" Prepared by: teacher of biology and chemistry of MBOU "Karpovskaya Secondary School" Chirkova Olga Aleksandrovna 2017
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Task 28. Problem on genetics. Genealogical method The genealogical method consists of analyzing pedigrees and allows us to determine the type of inheritance (dominant recessive, autosomal or sex-linked) of a trait, as well as its monogenic or polygenic nature. The person for whom a pedigree is being compiled is called a proband, and his proband's brothers and sisters are called sibs.
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Task 28. Problem on genetics. Genealogical method Symbols used in compiling genealogies
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Types of inheritance of traits Autosomal dominant type of inheritance. 1. Sick people occur in every generation. 2. Both men and women get sick 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 from 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 from healthy parents is 25% if they are heterozygous; 0% if both of them, 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-chromosome-linked (sex-linked) dominant type of inheritance 1. Patients occur in every generation. 2. Women are more affected. 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-chromosome (sex-linked) recessive type of inheritance. 1. Patients are not found in every generation. 2. Mostly men get sick. 3. The probability of having a sick boy born to healthy parents is 25%, and a sick girl is 0%.
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Types of inheritance of traits: Hollandic type of inheritance (Y-linked inheritance). 1. Sick people 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 having a sick boy from a sick father is 100%.
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Task 28. Problem on genetics. Genealogical method Stages of problem solving Determine the type of inheritance of a trait - dominant or recessive. Answer the questions: Is the trait found in all generations or not? Is the trait common among members of a pedigree? Are there cases of children being born with a trait if the parents do not exhibit this trait? Are there cases of children being born without the trait being studied if both parents have it? What part of the offspring carries the trait in families if one of the parents has it? 2. Determine whether the trait is inherited in a sex-linked manner. How often does the trait occur in both sexes (if it is rare, then which sex carries it more often)? persons of which sex inherit the trait from the father and mother who carry the trait? 3. Find out the formula for the splitting of descendants in one generation. And based on the analysis, determine the genotypes of all members of the pedigree.
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Task 28. Problem on genetics. Genealogical method Task 1. Using the pedigree presented in the figure, establish the nature of 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. Problem on genetics. Genealogical method Solution algorithm 1. Determine the type of inheritance: the trait occurs in all generations or not (the trait is dominant, because it is always passed on to the offspring) 2. Determine whether the trait is inherited in a gender-linked manner: more often in boys or girls (not linked to gender, since the trait is transmitted equally to sons and daughters). 3. Determine the genotypes of the parents: (woman aa (without the trait is homozygous), man Aa (with the trait) is heterozygous. 4. Solve the problem with genotypes: P: aa (f) x Aa (m. with the trait) G: a A a F1: Aa (m. with a sign), Aa (m. with a sign), aa (m. without a sign) P: Aa (m. 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 the offspring, not sex-linked since it is transmitted equally to both daughters and sons. Genotypes of parents: female: aa, male Aa (with the trait). 2) Genotypes of children in F1 women - Aa (with the trait) and aa, men - Aa (with the trait). 3) Genotypes of the descendants of F2 male - Aa (with the trait).
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Task 28. Problem on genetics. Genealogical method Task 2. Based on the pedigree shown in the figure, determine 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. Problem on 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 the trait), female aa (with the trait) 3. Solve the problem with genotypes: P: aa (f with a trait) x Aa (m. without a trait) G: a A a a F1: Aa (m. without a trait), Aa (f. without a trait) 4. Write down the answer : 1) The trait is recessive; 2) genotypes of the parents: mother - aa, father - AA or Aa; 3) genotypes of children: heterozygous son and daughter - Aa (allowed: other genetic symbolism that does not distort the meaning of solving the problem, indicating only one of the options for the father's genotype).
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Task 28. Problem on genetics. Genealogical method Task 3. Based on the pedigree shown in the figure, determine and explain the nature of inheritance of the trait (dominant or recessive, sex-linked or not) highlighted in black. Determine the genotypes of the descendants indicated in the diagram by numbers 3, 4, 8, 11 and explain the formation of their genotypes.
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Task 28. Problem on 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 linked to sex: more often in whom it occurs in boys or girls (linked with the X chromosome, since there is a generation skip). 3. Determine the genotypes of people indicated on the diagram by numbers 3, 4, 8, 11: 4. Write down the answer. 3 - female carrier - XHA 4 - male without the trait - XY 8 - male with the trait - XY 11 - female carrier - XHA
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Task 28. Problem on genetics. Genealogical method Task 4. Determine the type of inheritance, the genotype of the proband in the following pedigree Determination of the type of inheritance of the trait: The trait being studied is found only in males in each generation and is passed on from father to son (if the father is sick, then all sons also suffer from this disease), then we can think that the gene being studied 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 holandric 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 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 holandric inheritance. Proband genotype: XYa.
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Task 28. Problem on genetics. Codominance. Gene interaction. Problem 1. The cat color gene is linked to the X chromosome. Black coloring is determined by the XA gene, red coloring is determined 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. 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: P: cat ХХХB x cat ХБУ turtles. red G: XA XB XB Y.F1: red - XBY or XBXB sex-linked inheritance
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Task 28. Problem on genetics. Codominance. Gene interaction. Task 2. The genes for cat fur color are located on the X chromosome. Black coloring is determined by the gene XB red - Xb, heterozygotes have a tortoiseshell coloration. One tortoiseshell and one black kitten were born from a black cat and a ginger cat. Determine the genotypes of parents and offspring, the possible sex of kittens. Solution algorithm: Let's write down the crossing scheme: genotype of a black cat XB XB, genotype of a red cat - Xb Y, genotypes of kittens: tortoiseshell - XB Xb, Black - XB Y, gender of kittens: tortoiseshell - female, black - male.
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Task 28. Problem on genetics. Codominance. Gene interaction. 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 type 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 parents' blood groups. Indicate the possible genotypes and phenotypes (number) of the children's blood group. Make a diagram for solving the problem. Determine the probability of inheritance of blood group II in children. Solution algorithm: 1) parents have blood groups: group II - IAi0 (gametes IA, i0), group III - IВIВ (gametes IB); 2) possible phenotypes and genotypes of children’s blood groups: group IV (IAIB) and group III (IBi0); 3) the probability of inheriting blood group II is 0%.
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Task 28. Problem on genetics. Mono- and dihybrid crossing Problem 1. When crossing a corn plant with smooth colored seeds and a plant with wrinkled uncolored seeds, all the first generation hybrids had smooth colored seeds. From the analyzing crossing of F1 hybrids the following were obtained: 3800 plants with smooth colored seeds; 150 - with wrinkled colored ones; 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 diagram for solving the problem. Explain the formation of four phenotypic groups in an analytical cross.
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Task 28. Problem on genetics. Mono- and dihybrid crossing Solution algorithm: 1) First crossing: P AABB × aabb G AB × ab F1 AaBb 2) Analyzing crossing: 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. The other two phenotypic groups (149 and 150) are formed as a result of crossing over between allelic genes.
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Task 28. Problem on genetics. Mono- and dihybrid crossing Task 2. When crossing white guinea pigs with smooth hair with black guinea pigs with shaggy hair, the following 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% were white shaggy. Make a diagram of each cross. Determine the genotypes of parents and offspring. What is this crossing called and why is it carried out?
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Task 28. Problem on genetics. Mono- and dihybrid crossing Task 3. In common peas, the pink color of the corolla dominates over the white one, and the tall stem dominates over the dwarf one. When crossing a plant with a tall stem and pink flowers with a plant with pink flowers and a dwarf stem, 63 plants with a tall stem and pink flowers were obtained, 58 with pink flowers and a dwarf stem, 18 with white flowers and a tall stem, 20 with white flowers and dwarf stem. Make a diagram for solving the problem. Determine the genotypes of the original plants and descendants. Explain the nature of inheritance of traits and the formation of four phenotypic groups.
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Task 28. Problem on 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, therefore the inheritance of traits is independent.
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Task 28. Problem on genetics. Gene linkage Problem 1. A married couple in which both spouses had normal vision gave birth to 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 ♂ ХD, У; ♀ 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 is XDY, and the other is XdY.
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Task 28. Problem on genetics. Gene linkage Problem 2. A woman, a carrier of the recessive hemophilia gene, married a healthy man. Determine the genotypes of the parents, and the ratio of genotypes and phenotypes in the expected offspring. Solution algorithm 1) Genotypes of parents XH Xh and XHY; 2) genotypes of the offspring - ХН Хh, ХН ХН, ХН У, ХhУ; The genotype ratio is 1:1:1:1 3) the daughters are a carrier of the hemophilia gene, healthy, and the sons are healthy and have hemophilia. Phenotype ratio 2 (healthy girls): 1 (healthy boy) : 1 (hemophilic boy)
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Task 28. Problem on genetics. Gene linkage Task 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 the marriage of a dihomozygous woman, normal for both alleles, and an albino man with hemophilia. Make a diagram 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 who is outwardly normal for both alleles, but is a carrier of the genes for albinism and hemophilia; A boy who is outwardly normal for both alleles, but is a carrier of the albinism gene.
For this task you can get 3 points on the Unified State Exam in 2020
The topic of assignment 28 of the Unified State Exam in biology was “Supraorganismal systems and the evolution of the world.” Many schoolchildren note the difficulty of this test due to the large volume it covers. educational material, and also due to the construction of the ticket. In task No. 28, the compiler is the Russian FIPI, Federal Institute pedagogical measurements, offers six answer options for each question, the correct ones can 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 features out of the six listed are characteristic of animal cells”), but in most cases the student must decide for himself the number of answer options he chooses as correct.
Questions on task 28 of the Unified State Exam in biology may also touch on the basics of biology. Be sure to repeat before the exams - what is the absence of artificial and natural ecosystems, aquatic and terrestrial, meadow and field, what does the rule of the ecological pyramid sound like and where it is applicable, what is biogeocenosis and agrocenosis. Some questions are logical in nature, you need to not only rely on theory from school textbook, but also to think logically: “In a mixed forest, plants are arranged in tiers, and this is the reason for the decrease in competition between the birch and another living organism. Which one?” The following answers include cockchafer, bird cherry, mushrooms, rose hips, hazel, and mice. In this case, the student must remember that competition always occurs for the same resources, in in this case(with a tiered arrangement of plants) - for light, so from the list you need to select only trees and shrubs - bird cherry, rose hips and hazel.
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