Biology genetics assignments. Assignments on genetics for the Unified State Exam in biology
with examples
explanations and solutions
To solve problems in genetics you need:
introduce symbols of signs. A dominant gene is indicated by a capital letter, and a recessive gene is indicated by a capital letter;
find out what kind of crossing takes place in this case (mono-, di- or polyhybrid);
write down the solution to the problem schematically and draw a conclusion about the probability of the requested event in % or fractions of unity.
Monohybrid crossing.
One pair of alternative (mutually exclusive) traits is taken into account, therefore, one pair of allelic genes.
Task No. 1
In tomatoes, the gene that produces red fruits is dominant over the gene that produces yellow fruits. What color fruits will be produced by plants obtained from crossing heterozygous red-fruited plants with yellow-fruited plants? What are their genotypes?
Solution to problem No. 1
The gene responsible for the red color of fruits is dominant, we will denote it as A, the gene for the yellow color of fruits is recessive, we will denote it as a. The problem of monohybrid crossing, because in the condition, the plants differ in 1 pair of mutually exclusive characters (fruit color). One of the parental individuals is heterozygous, therefore, this individual carries the genes of one allelic pair in a different state, one gene is dominant, the other is recessive (Aa). Such an individual produces two types of gametes (A, a). The second individual is yellow-fruited, both genes are in the same state (aa), which means that the individual is homozygous and produces one type of gametes (a).
Knowing the genotypes of the parents, we will write down the solution to the problem and answer the question posed.
A – red fruits
a – yellow fruits
R – parents
G – gametes
F 1 – first generation
x – crossing sign
R Aa x aa
red yellow
red yellow
Answer: F 1 hybrids demonstrate phenotypic cleavage in a 1:1 ratio - 50% red-fruited, 50% yellow-fruited tomatoes. According to the genotype, splitting in a ratio of 1:1 - 50% heterozygous individuals (Aa), 50% homozygous individuals (aa).
Problems for independent solution.
3Problem No. 2.
In Drosophila, gray body color is dominant over black. By crossing flies with a gray body and flies with a black body, F1 hybrids were obtained, which subsequently, when crossed with each other, produced 192 individuals of the next generation.
1. How many types of gametes are formed in the F I hybrid?
2. How many different phenotypes are there among the descendants of F 2?
3. How many different genotypes are there among the descendants of F 2?
4. How many homozygous gray flies are there in F 2 (theoretically)?
5. How many blacks are there in F 2 (theoretically)?
Task No. 3.
In humans, the gene for long eyelashes is dominant over the gene for short eyelashes. A woman with long eyelashes, whose father had short eyelashes, marries a man with short eyelashes.
1. How many types of gametes are produced in a man?
2. How many types of gametes does a woman produce?
3. How many different genotypes can there be among the children of this married couple?
4. What is the probability that a child in this family will be born with long eyelashes?
5. What is the probability that a child in this family will be born with short eyelashes?
Task No. 4.
In humans, the gene for the early development of hypertension dominates over the gene that determines the normal development of the trait. In the family, both spouses suffer from early-onset hypertension; their only daughter has normal blood pressure. She is married and has two children. One of the daughter's children has normal blood pressure, while the other developed hypertension early on.
1. How many different genotypes can there be among the children of the above spouses?
2. How many types of gametes does the daughter produce?
3. How many types of gametes does the daughter’s husband produce?
4. What is the probability that the daughter of these spouses will have a child with hypertension?
5. How many different genotypes can there be among the daughter’s grandchildren?
The study of the basic laws of heredity and variability of organisms is one of the most complex, but very promising tasks facing modern natural science. In this article we will consider both the basic theoretical concepts and postulates of science, and we will understand how to solve problems in genetics.
Relevance of studying patterns of heredity
Two of the most important branches of modern science - medicine and breeding - are developing thanks to the research of genetic scientists. The biological discipline itself, the name of which was proposed in 1906 by the English scientist W. Betson, is not so much theoretical as practical. Anyone who decides to seriously understand the mechanism of inheritance of various traits (for example, eye color, hair, blood type) will first have to study the laws of heredity and variability, and also figure out how to solve problems in human genetics. This is precisely the issue we will address.
Basic concepts and terms
Each industry has a specific, unique set of basic definitions. If we are talking about science that studies the processes of transmission of hereditary characteristics, we will understand the following terms by the latter: gene, genotype, phenotype, parental individuals, hybrids, gametes, and so on. We will meet each of them when we study the rules that explain to us how to solve biology problems involving genetics. But first we will study the hybridological method. After all, it is precisely this that underlies genetic research. It was proposed by the Czech naturalist G. Mendel in the 19th century.
How are traits inherited?
The patterns of transmission of the properties of an organism were discovered by Mendel thanks to the experiments he conducted with a well-known plant - The hybridological method is the crossing of two units that differ from each other in one pair of characteristics (monohybrid crossing). If the experiment involves organisms that have several pairs of alternative (opposite) traits, then they speak of polyhybrid crossing. The scientist proposed the following form of recording the progress of hybridization of two pea plants that differ in seed color. A - yellow paint, and - green.
In this entry, F1 are first (I) generation hybrids. They are all absolutely uniform (the same), since they contain A, which controls the yellow color of the seeds. The above entry corresponds to the first (F1 Hybrid Uniformity Rule). Knowledge of it explains to students how to solve problems in genetics. Grade 9 has a biology program in which the hybridological method of genetic research is studied in detail. It also discusses Mendel's second (II) rule, called the law of splitting. According to it, in F2 hybrids obtained from crossing two first-generation hybrids with each other, a splitting ratio is observed in the ratio of phenotype 3 to 1, and in genotype 1 to 2 and 1.
Using the above formulas, you will understand how to solve problems in genetics without errors, if in their conditions you can apply Mendel’s first or already known II law, taking into account that crossing occurs with one of the genes.
Law of independent combination of feature states
If the parents differ in two pairs of alternative traits, such as seed color and shape, in plants such as the common pea, then a Pinnett grid should be used in the genetic cross.
Absolutely all hybrids that are the first generation obey Mendel’s rule of uniformity. That is, they are yellow, with a smooth surface. Continuing to cross F1 plants with each other, we will obtain second generation hybrids. To figure out how to solve problems in genetics, 10th grade biology lessons use a dihybrid cross recording using the phenotypic segregation formula 9:3:3:1. Provided that the genes are located in different pairs, Mendel's third postulate can be used - the law of independent combinations of character states.
How are blood types inherited?
The mechanism of transmission of such a trait as a person’s blood type does not correspond to the patterns we discussed earlier. That is, it does not obey Mendel's first and second laws. This is explained by the fact that such a trait as blood type, according to Landsteiner’s research, is controlled by three alleles of gene I: A, B and 0. Accordingly, the genotypes will be as follows:
- The first group is 00.
- The second is AA or A0.
- The third group is BB or B0.
- The fourth is AB.
Gene 0 is a recessive allele to genes A and B. And the fourth group is the result of codominance (mutual presence of genes A and B). It is this rule that must be taken into account in order to know how to solve problems in genetics regarding blood groups. But that is not all. To establish the genotypes of children by blood group born from parents with different blood groups, we will use the table below.
Morgan's theory of heredity
Let's return to the section of our article “The Law of Independent Combination of Character States,” in which we looked at how to solve problems in genetics. like Mendel’s III law itself, to which it obeys, is applicable for allelic genes located in the homologous chromosomes of each pair.
In the mid-20th century, the American geneticist T. Morgan proved that most traits are controlled by genes that are located on the same chromosome. They have a linear arrangement and form clutch groups. And their number is exactly equal to the haploid set of chromosomes. During the process of meiosis, leading to the formation of gametes, not individual genes, as Mendel believed, but entire complexes of them, called linkage groups by Morgan, enter the germ cells.
Crossing over
During prophase I (also called the first division of meiosis), sections (lucuses) are exchanged between the internal chromatids of homologous chromosomes. This phenomenon is called crossing over. It underlies hereditary variability. Crossing over is especially important for studying branches of biology that study hereditary human diseases. Using the postulates set forth in Morgan's chromosomal theory of heredity, we will define an algorithm that answers the question of how to solve problems in genetics.
Sex-linked inheritance cases are a special case of transmission of genes that are located on the same chromosome. The distance that exists between genes in linkage groups is expressed as a percentage - morganids. And the strength of adhesion between these genes is directly proportional to the distance. Therefore, crossing over most often occurs between genes that are located far from each other. Let us consider the phenomenon of linked inheritance in more detail. But first, let us remember which elements of heredity are responsible for the sexual characteristics of organisms.
Sex chromosomes
In the human karyotype, they have a specific structure: in females they are represented by two identical X chromosomes, and in males in a sexual pair, in addition to the X chromosome, there is also a Y variant, which differs both in shape and in the set of genes. This means that it is not homologous to the X chromosome. Hereditary human diseases such as hemophilia and color blindness arise due to the “breakdown” of individual genes on the X chromosome. For example, from the marriage of a hemophilia carrier with a healthy man, the birth of such offspring is possible.
The above course of genetic crossing confirms the fact that the gene that controls blood clotting is linked to the sex X chromosome. This scientific information is used to teach students techniques that determine how to solve problems in genetics. The 11th grade has a biology program, which covers in detail such sections as “genetics”, “medicine” and “human genetics”. They allow students to study hereditary human diseases and know the reasons why they occur.
Gene interaction
The transmission of hereditary characteristics is a rather complex process. The above diagrams become understandable only if students have a basic minimum of knowledge. It is necessary because it provides mechanisms that answer the question of how to learn to solve problems in biology. Genetics studies the forms of interaction between genes. These are polymerization, epistasis, complementarity. Let's talk about them in more detail.
The example of the inheritance of hearing in humans is an illustration of this type of interaction as complementarity. Hearing is controlled by two pairs of different genes. The first is responsible for the normal development of the cochlea of the inner ear, and the second is responsible for the functioning of the auditory nerve. Marriages of deaf parents, each of whom is a recessive homozygote for each of two pairs of genes, produce children with normal hearing. Their genotype contains both dominant genes that control the normal development of the hearing system.
Pleiotropy
This is an interesting case of gene interaction, in which the phenotypic manifestation of several traits depends on one gene present in the genotype. For example, human populations of some representatives have been discovered in western Pakistan. They lack sweat glands in certain areas of the body. At the same time, such people were diagnosed with the absence of some molars. They could not form in the process of ontogenesis.
Animals such as Karakul sheep have a dominant W gene, which controls both fur color and normal stomach development. Let's consider how the W gene is inherited when two heterozygous individuals are crossed. It turns out that in their offspring, ¼ of the lambs with the WW genotype die due to abnormalities in the development of the stomach. Moreover, ½ (those with gray fur) are heterozygous and viable, and ¼ are individuals with black fur and normal stomach development (their WW genotype).
Genotype is a complete system
The multiple action of genes, polyhybrid crossing, and the phenomenon of linked inheritance serve as indisputable proof of the fact that the totality of genes in our body is an integral system, although it is represented by individual gene alleles. They can be inherited according to Mendel's laws, independently or by loci, linked together, obeying the postulates of Morgan's theory. Considering the rules responsible for how to solve problems in genetics, we were convinced that the phenotype of any organism is formed under the influence of both allelic and developmental traits of one or more traits.
Among the tasks on genetics on the Unified State Exam in biology, 6 main types can be distinguished. The first two - to determine the number of gamete types and monohybrid crossing - are most often found in part A of the exam (questions A7, A8 and A30).
Problems of types 3, 4 and 5 are devoted to dihybrid crossing, inheritance of blood groups and sex-linked traits. Such tasks make up the majority of C6 questions in the Unified State Exam.
The sixth type of task is mixed. They consider the inheritance of two pairs of traits: one pair is linked to the X chromosome (or determines human blood groups), and the genes of the second pair of traits are located on autosomes. This class of tasks is considered the most difficult for applicants.
This article outlines theoretical foundations of genetics necessary for successful preparation for task C6, as well as solutions to problems of all types are considered and examples are given for independent work.
Basic terms of genetics
Gene- this is a section of a DNA molecule that carries information about the primary structure of one protein. A gene is a structural and functional unit of heredity.
Allelic genes (alleles)- different variants of one gene, encoding an alternative manifestation of the same trait. Alternative signs are signs that cannot be present in the body at the same time.
Homozygous organism- an organism that does not split according to one or another characteristic. Its allelic genes equally influence the development of this trait.
Heterozygous organism- an organism that produces splitting according to certain characteristics. Its allelic genes have different effects on the development of this trait.
Dominant gene is responsible for the development of a trait that manifests itself in a heterozygous organism.
Recessive gene is responsible for a trait whose development is suppressed by a dominant gene. A recessive trait occurs in a homozygous organism containing two recessive genes.
Genotype- a set of genes in the diploid set of an organism. The set of genes in a haploid set of chromosomes is called genome.
Phenotype- the totality of all the characteristics of an organism.
G. Mendel's laws
Mendel's first law - the law of hybrid uniformity
This law was derived based on the results of monohybrid crosses. For the experiments, two varieties of peas were taken, differing from each other in one pair of characteristics - the color of the seeds: one variety was yellow in color, the second was green. The crossed plants were homozygous.
To record the results of crossing, Mendel proposed the following scheme:
Yellow color of seeds 
- green color of seeds
| (parents) | ||
| (gametes) |
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| (first generation) | (all plants had yellow seeds) |
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Statement of the law: when crossing organisms that differ in one pair of alternative characteristics, the first generation is uniform in phenotype and genotype.
Mendel's second law - the law of segregation
Plants were grown from seeds obtained by crossing a homozygous plant with yellow colored seeds with a plant with green colored seeds and obtained by self-pollination.
(plants have a dominant trait - recessive) |
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Statement of the law: in the offspring obtained from crossing first-generation hybrids, there is a split in phenotype in the ratio , and in genotype -.
Mendel's third law - the law of independent inheritance
This law was derived from data obtained from dihybrid crosses. Mendel considered the inheritance of two pairs of characteristics in peas: color and seed shape.
As parental forms, Mendel used plants homozygous for both pairs of traits: one variety had yellow seeds with smooth skin, the other had green and wrinkled seeds.
Yellow color of seeds, - green color of seeds,
- smooth form, - wrinkled form.
(yellow smooth). |
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Mendel then grew plants from seeds and obtained second-generation hybrids through self-pollination.
The Punnett grid is used to record and determine genotypes
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There was a split into phenotypic classes in the ratio. All seeds had both dominant traits (yellow and smooth), - the first dominant and second recessive (yellow and wrinkled), - the first recessive and second dominant (green and smooth), - both recessive traits (green and wrinkled).
When analyzing the inheritance of each pair of traits, the following results are obtained. In parts of yellow seeds and parts of green seeds, i.e. ratio . Exactly the same ratio will be for the second pair of characteristics (seed shape).
Statement of the law: when crossing organisms that differ from each other in two or more pairs of alternative traits, genes and their corresponding traits are inherited independently of each other and combined in all possible combinations.
Mendel's third law is true only if the genes are located in different pairs of homologous chromosomes.
Law (hypothesis) of “purity” of gametes
When analyzing the characteristics of hybrids of the first and second generations, Mendel established that the recessive gene does not disappear and does not mix with the dominant one. Both genes are expressed, which is only possible if the hybrids form two types of gametes: some carry a dominant gene, others carry a recessive one. This phenomenon is called the gamete purity hypothesis: each gamete carries only one gene from each allelic pair. The hypothesis of gamete purity was proven after studying the processes occurring in meiosis.
The hypothesis of the "purity" of gametes is the cytological basis of Mendel's first and second laws. With its help, it is possible to explain the splitting by phenotype and genotype.
Analysis cross
This method was proposed by Mendel to determine the genotypes of organisms with a dominant trait that have the same phenotype. To do this, they were crossed with homozygous recessive forms.
If, as a result of crossing, the entire generation turned out to be the same and similar to the analyzed organism, then one could conclude: the original organism is homozygous for the trait being studied.
If, as a result of crossing, a split in the ratio was observed in a generation, then the original organism contains genes in a heterozygous state.
Inheritance of blood groups (AB0 system)
Inheritance of blood groups in this system is an example of multiple allelism (the existence of more than two alleles of one gene in a species). In the human population, there are three genes encoding red blood cell antigen proteins that determine people's blood types. The genotype of each person contains only two genes that determine his blood type: group one; second and ; third and fourth.
Inheritance of sex-linked traits
In most organisms, sex is determined during fertilization and depends on the number of chromosomes. This method is called chromosomal sex determination. Organisms with this type of sex determination have autosomes and sex chromosomes - and.
In mammals (including humans), the female sex has a set of sex chromosomes, while the male sex has a set of sex chromosomes. The female sex is called homogametic (forms one type of gametes); and the male one is heterogametic (forms two types of gametes). In birds and butterflies, the homogametic sex is male, and the heterogametic sex is female.
The Unified State Exam includes tasks only for traits linked to the - chromosome. They mainly concern two human characteristics: blood clotting (- normal; - hemophilia), color vision (- normal, - color blindness). Tasks on the inheritance of sex-linked traits in birds are much less common.
In humans, the female sex can be homozygous or heterozygous for these genes. Let's consider possible genetic sets in a woman using hemophilia as an example (a similar picture is observed with color blindness): - healthy; - healthy, but is a carrier; - sick. The male sex is homozygous for these genes, because -chromosome does not have alleles of these genes: - healthy; - is ill. Therefore, most often men suffer from these diseases, and women are their carriers.
Typical USE tasks in genetics
Determination of the number of gamete types
The number of gamete types is determined using the formula: , where is the number of gene pairs in the heterozygous state. For example, an organism with a genotype does not have genes in a heterozygous state, i.e. , therefore, and it forms one type of gametes. An organism with a genotype has one pair of genes in a heterozygous state, i.e. , therefore, and it forms two types of gametes. An organism with a genotype has three pairs of genes in a heterozygous state, i.e. , therefore, and it forms eight types of gametes.
Mono- and dihybrid crossing problems
For monohybrid crossing
Task: Crossed white rabbits with black rabbits (black color is the dominant trait). In white and black. Determine the genotypes of parents and offspring.
Solution: Since segregation according to the studied trait is observed in the offspring, therefore, the parent with the dominant trait is heterozygous.
| (black) | (white) | |
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| (black) : (white) |
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For dihybrid crossing
Dominant genes are known
Task: Crossed normal-sized tomatoes with red fruits with dwarf tomatoes with red fruits. All plants were of normal growth; - with red fruits and - with yellow ones. Determine the genotypes of parents and offspring if it is known that in tomatoes, red fruit color dominates yellow, and normal growth dominates dwarfism.
Solution: Let's designate dominant and recessive genes: - normal growth, - dwarfism; - red fruits, - yellow fruits.
Let's analyze the inheritance of each trait separately. All descendants have normal growth, i.e. no segregation for this trait is observed, therefore the initial forms are homozygous. Segregation is observed in fruit color, so the original forms are heterozygous.
(dwarfs, red fruits) |
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| (normal growth, red fruits) (normal growth, red fruits) (normal growth, red fruits) (normal growth, yellow fruits) |
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Dominant genes unknown
Task: Two varieties of phlox were crossed: one has red saucer-shaped flowers, the second has red funnel-shaped flowers. The offspring produced were red saucer, red funnel, white saucer and white funnel. Determine the dominant genes and genotypes of the parental forms, as well as their descendants.
Solution: Let's analyze the splitting for each characteristic separately. Among the descendants of plants with red flowers are, with white flowers -, i.e. . That's why it's red, - white color, and the parental forms are heterozygous for this trait (since there is cleavage in the offspring).
There is also a split in flower shape: half of the offspring have saucer-shaped flowers, the other half have funnel-shaped flowers. Based on these data, it is not possible to unambiguously determine the dominant trait. Therefore, we accept that - saucer-shaped flowers, - funnel-shaped flowers.
(red flowers, saucer-shaped) |
(red flowers, funnel-shaped) |
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Red saucer-shaped flowers,
- red funnel-shaped flowers,
- white saucer-shaped flowers,
- white funnel-shaped flowers.
Solving problems on blood groups (AB0 system)
Task: the mother has the second blood group (she is heterozygous), the father has the fourth. What blood types are possible in children?
Solution:
(the probability of having a child with the second blood group is , with the third - , with the fourth - ). |
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Solving problems on the inheritance of sex-linked traits
Such tasks may well appear in both Part A and Part C of the Unified State Examination.
Task: a carrier of hemophilia married a healthy man. What kind of children can be born?
Solution:
| girl, healthy () girl, healthy, carrier () boy, healthy () boy with hemophilia () |
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Solving problems of mixed type
Task: A man with brown eyes and a blood type married a woman with brown eyes and a blood type. They had a blue-eyed child with a blood type. Determine the genotypes of all individuals indicated in the problem.
Solution: Brown eye color dominates blue, therefore - brown eyes, - Blue eyes. The child has blue eyes, so his father and mother are heterozygous for this trait. The third blood group can have a genotype or, the first - only. Since the child has the first blood group, therefore, he received the gene from both his father and mother, therefore his father has the genotype.
| (father) | (mother) | |
| (was born) | ||
Task: A man is colorblind, right-handed (his mother was left-handed) married to a woman with normal vision (her father and mother were completely healthy), left-handed. What kind of children can this couple have?
Solution: In a person, better control of the right hand dominates over left-handedness, therefore - right-handed, - left-handed. The genotype of the man (since he received the gene from a left-handed mother), and women - .
A colorblind man has the genotype, and his wife has the genotype, because. her parents were completely healthy.
| R | ||
| right-handed girl, healthy, carrier () left-handed girl, healthy, carrier () right-handed boy, healthy () left-handed boy, healthy () |
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Problems to solve independently
- Determine the number of gamete types in an organism with genotype.
- Determine the number of gamete types in an organism with genotype.
- Crossed tall plants with short plants. B - all plants are medium in size. What will it be?
- Crossed a white rabbit with a black rabbit. All rabbits are black. What will it be?
- Two rabbits with gray fur were crossed. In with black wool, - with gray and with white. Determine the genotypes and explain this segregation.
- A black hornless bull was crossed with a white horned cow. We got black hornless, black horned, white horned and white hornless. Explain this split if black color and lack of horns are dominant characteristics.
- Drosophila flies with red eyes and normal wings were crossed with fruit flies with white eyes and defective wings. The offspring are all flies with red eyes and defective wings. What will be the offspring from crossing these flies with both parents?
- A blue-eyed brunette married a brown-eyed blonde. What kind of children can be born if both parents are heterozygous?
- A right-handed man with a positive Rh factor married a left-handed woman with a negative Rh factor. What kind of children can be born if a man is heterozygous only for the second characteristic?
- The mother and father have the same blood type (both parents are heterozygous). What blood type is possible in children?
- The mother has a blood type, the child has a blood type. What blood type is impossible for a father?
- The father has the first blood group, the mother has the second. What is the probability of having a child with the first blood group?
- A blue-eyed woman with a blood type (her parents had a third blood group) married a brown-eyed man with a blood type (his father had blue eyes and a first blood group). What kind of children can be born?
- A hemophilic man, right-handed (his mother was left-handed) married a left-handed woman with normal blood (her father and mother were healthy). What children can be born from this marriage?
- Strawberry plants with red fruits and long-petioled leaves were crossed with strawberry plants with white fruits and short-petioled leaves. What kind of offspring can there be if red color and short-petioled leaves dominate, while both parent plants are heterozygous?
- A man with brown eyes and a blood type married a woman with brown eyes and a blood type. They had a blue-eyed child with a blood type. Determine the genotypes of all individuals indicated in the problem.
- Melons with white oval fruits were crossed with plants that had white spherical fruits. The offspring produced the following plants: with white oval, white spherical, yellow oval and yellow spherical fruits. Determine the genotypes of the original plants and descendants, if in a melon the white color dominates over the yellow, the oval shape of the fruit dominates over the spherical.
Answers
- type of gametes.
- types of gametes.
- type of gametes.
- high, medium and low (incomplete dominance).
- black and white.
- - black, - white, - gray. Incomplete dominance.
- Bull: , cow - . Offspring: (black hornless), (black horned), (white horned), (white hornless).
- - Red eyes, - white eyes; - defective wings, - normal. Initial forms - and, offspring.
Crossing results:
A) - - Brown eyes, - blue; - dark hair, - blond. Father mother - .
- brown eyes, dark hair
- brown eyes, blond hair
- blue eyes, dark hair
- blue eyes, blond hair - - right-handed, - left-handed; - Rh positive, - Rh negative. Father mother - . Children: (right-handed, Rh positive) and (right-handed, Rh negative).
- Father and mother - . Children may have a third blood group (probability of birth - ) or first blood group (probability of birth - ).
- Mother, child; he received the gene from his mother, and from his father - . The following blood groups are impossible for the father: second, third, first, fourth.
- A child with the first blood group can only be born if his mother is heterozygous. In this case, the probability of birth is .
- - Brown eyes, - blue. Female Male . Children: (brown eyes, fourth group), (brown eyes, third group), (blue eyes, fourth group), (blue eyes, third group).
- - 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).
- - red fruits, - white; - short-petioled, - long-petioled.
Parents: and. Offspring: (red fruits, short-petioled), (red fruits, long-petioled), (white fruits, short-petioled), (white fruits, long-petioled).
Strawberry plants with red fruits and long-petioled leaves were crossed with strawberry plants with white fruits and short-petioled leaves. What kind of offspring can there be if red color and short-petioled leaves dominate, while both parent plants are heterozygous? - - Brown eyes, - blue. Female Male . Child:
- - white color, - yellow; - oval fruits, - round. Source plants: and. Offspring:
with white oval fruits,
with white spherical fruits,
with yellow oval fruits,
with yellow spherical fruits.
Solving typical problems in genetics
To solve genetic problems, you should use the proposed algorithm:
Determine the number of features that are analyzed;
Determine the variants of manifestation of the named characteristics (dominant, recessive, intermediate);
Determine the type of crossing;
Analyze the phenotypes described in the problem statement and record the genotypes of the parental individuals;
Identify and record possible crossbreeding options;
Recreate the crossing pattern;
Write down your answer.
Sample solution to a problem in genetics
Task. In humans, brown eye color is dominant over blue. A homozygous brown-eyed man marries a blue-eyed woman. What eye color will their children have?
Answer: all children will have brown eyes.
Tasks by genetics
- In a person, the ability to use predominantly the right hand is a dominant trait, and the left hand is a recessive trait. A right-handed man, whose mother was left-handed, married a right-handed woman who had 2 sisters, one of whom was left-handed. What is the probability of having a left-handed child?
- In the Drosophila fly, gray body color dominates over black. When gray flies were crossed, the offspring produced 1,390 gray individuals and 460 black individuals. Determine the genotypes of the parental individuals.
- A blue-eyed, right-handed man whose father was left-handed married a brown-eyed, left-handed woman from a family whose members had all had brown eyes for generations. What kind of children can they have?
- In a guinea pig, wavy fur dominates over smooth fur. Write down the genotypes of all animals in such crosses: a) with wavy hair × with smooth hair = all offspring with wavy hair; b) with wavy coat × with smooth coat = half of the offspring with wavy coat, half with smooth coat; c) with smooth coat × with smooth coat = all offspring are smooth coated.
Complex problems in genetics.
- In humans, brown eyes are a dominant trait, blue eyes are a recessive trait. A blue-eyed man whose parents had brown eyes married a brown-eyed woman whose father had blue eyes and whose mother had brown eyes. What kind of eyes might their children have? Determine the genotypes of all mentioned individuals.
- A long-haired black male guinea pig was crossed with a black short-haired female. We obtained 15 pigs with short black hair, 13 with long black hair, 4 with short white hair, and 5 with long white hair. Determine the genotypes of the parents if black and long hair are the dominant traits.
Problems in genetics - Deviations during splitting from typical quantitative relationships .
Training tasks
- What kind of children can be from the marriage of a man and a woman with wavy hair, if curly hair is a dominant trait, and straight hair is recessive?
- The mother has blood group I, and the father has blood group IV. Can children inherit the blood type of one of their parents?
- The offspring from crossing a gray Drosophila with a black Drosophila resulted in 290 black and 286 gray Drosophila. What are the genotypes of their descendants?
- In a guinea pig, curly hair and black coloring are dominant traits, while smooth fur and white coloring are recessive. As a result of crossing a black curly pig with a white smooth-haired pig, 9 curly black pigs and 11 curly white pigs were obtained in the offspring. Determine the genotypes of the parents.
- Foxes with the PP genotype are platinum in color, while PP are silver. The dominant P allele is lethal. What will be the offspring of a cross between a platinum fox and a silver fox?
- When black mice are crossed with each other, black offspring always appear. When yellow mice are crossed with each other, one third of the offspring turns out to be black, and two thirds are yellow. How can this be explained?
- Parents with blood groups II and III gave birth to a child with blood group I, sick with sickle cell anemia (autosomal inheritance unlinked to blood groups with incomplete dominance). Determine the probability of birth: a) sick children with blood group I; b) sick children with blood group II; c) sick children with blood group IV.
Problems in genetics on the topic Chromosomal theory of heredity
Solutions tasks
- According to hybridological analysis, the phenotypic traits of body color and wing structure in Drosophila are inherited in a linked manner, however, this linkage is not absolute. The gray color of the body dominates over the black, and the normal structure of the wings dominates over the rudimentary wings. As a result of crossing diheterozygous females with males that have a black body and underdeveloped wings, the following flies were obtained in the offspring: 1,394 - gray with normal wings, 1,418 - black with rudimentary wings, 288 - gray with rudimentary wings and 287 - black with normal wings. Determine the genotype of parents and offspring, as well as the distance between genes.
- According to hybridological analysis, the K, L and M genes are contained in the same linkage group. Crossing over between the K and L genes occurs with a frequency of 8%, and between the K and M genes with a frequency of 11%. Construct a genetic map of this chromosome.
Creative tasks in genetics
In a tomato, the tall stem (A) dominates the dwarf one, and the round fruits (B) dominate the pear-shaped ones. The genes that determine the height of the stem and the shape of the fruit are linked and are contained in the chromosome at a distance of 10 morganids. Homozygous tall plants with round fruits crossed with dwarf plants that have pear-shaped fruits. What genotypes of parents, hybrids and what types of gametes and in what percentage form the first generation hybrids?
Objectives - Genetics of sex. Sex-linked inheritance
Sex-linked inheritance of traits
- In cats, red fur color is dominant over black. Heterozygotes have a tortoiseshell coloration. Sex-linked color gene (X chromosome). What kind of descendants can there be if: a) the cat is black and the cat is red; b) the cat is black and the cat is tortoiseshell; c) the cat is red and the cat is black?
- Color blindness is a recessive trait, the gene for which is located on the X chromosome. What kind of children can be expected from marriage: a) a man is the norm, a woman is a carrier; b) man- colorblind, female - norm; c) a man is color blind, a woman is a carrier.
CREATIVE TASK
A woman whose mother had color blindness and whose father had hemophilia, married to a man who has both diseases. Determine the probability of having children in this family who will also have both of these diseases.
Tasks - Gene interaction
Solving problems of gene interaction.
- In budgerigars, allele A determines the yellow color of the feather, B - blue, when A and B interact, the color is green, an individual with the aabb genotype - white. By crossing heterozygous individuals with yellow and blue feathers, 20 parrots were obtained. How many of them are white?
- When crossing white-fruited and green-fruited pumpkins in the first generation, 50% white-fruited and 50% yellow-fruited pumpkins were obtained. Determine the genotypes of parents and hybrids.
- Human growth is controlled by several pairs of unlinked genes that interact based on the type of polymer. If we neglect environmental factors and conditionally limit ourselves to only three pairs of genes, then we can assume that in some population the shortest people have all recessive genes and a height of 150 cm, and the tallest people have all dominant genes and a height of 180 cm. Determine the height of people who are heterozygous behind all three pairs of growth genes.
Complex tasks
In humans, one form of hereditary deafness is determined by two recessive alleles of different genes. For normal hearing, the presence of two dominant alleles is necessary, one of which determines the development of the cochlea, and the second - the auditory nerve. In the family, the parents are deaf, and their two children have normal hearing. Determine the type of gene interaction and genotypes of family members.
Tasks - Types of mutations
Solutions tasks And exercises
Task 1. In one animal population, the number of chromosomal mutations is close to 7 per 1,000 births. Calculate the frequency of chromosomal mutations in this population. (The frequency of mutations can be calculated using the formula: Pm = M / 2N, where Pm is the frequency of occurrence of mutations, M is the number of identified mutant phenotypes, N is the total number of examined organisms).
Task 2. Fill in the table of changes in the number of chromosomes in different organisms as a result of mutational variation.
Determine the type and type of mutations: ________________
Task 3. The chromosome has the following sequence of linearly arranged genes: ABCDEFHMNK. After certain mutations, the order of genes can change. Fill out the table and determine the type and type of mutations associated with the rearrangement of this chromosome.
Task 4. In the wild type allele (original gene) - CCC GGT ACC CCC GGG - the following mutation took place: CAC GGT ACC CCC GTG. Determine the type of mutation. Compare fragments of a protein molecule that is encoded by the original and mutant genes.
Independent solution tasks
Task 1. As a result of a mutation in the gene region - HGC TTG CAC ACT AGG CAA - a substitution took place in the third triplet - cytosine was identified instead of adenine. Write down the amino acid composition of the polypeptide before and after the mutation.
Problem 2. Which change in the coding DNA strand - AGG TGA CTC ACG ATT - will have a greater impact on the primary structure of the protein: the loss of one first nucleotide from the second triplet or the loss of the entire second triplet? Write down the corresponding sections of protein molecules normally and after mutational changes in the gene.
Problems from population genetics
Solutions tasks
Task 1. Calculate the frequency of dominant and recessive alleles in a group consisting of 160 individuals with the CC genotype and 40 individuals with the CC genotype ss.
Task 2. The population includes 9% AA homozygotes, 49% aa homozygotes and 42% Aa heterozygotes. Determine the frequency of alleles A and a in the population.
Task 3. In the fox population in a certain area, 9,991 red individuals and 9 albino foxes were identified. Albinism is encoded by a recessive gene, and red coloration is encoded by its dominant allele. Determine the genetic structure of this fox population, assuming it is ideal. How many homozygous red foxes are there in this population?
Independent solution tasks
Task 1. In Europe, there is 1 albino for every 20 thousand people. What percentage of individuals are heterozygous carriers of the albinism allele?
Task 2. In the population of mongrel dogs, 2,457 short-legged animals and 243 with normal legs were identified. Short legs in dogs are a dominant trait, and normal leg length is recessive. Based on the Hardy-Weinberg law, determine: a) the frequency of occurrence of dominant and recessive alleles (in%); b) the percentage of short-legged dogs that, when crossed with each other, would never produce puppies with normal endings.
In your school biology course, in high school, you probably became acquainted, or will still become acquainted with genetic problems. Genetics is an extremely fascinating science. She comprehends the validity of variability and heredity. Representatives of each biological species reproduce those similar to themselves. However, no individuals are identical; all descendants differ to a greater or lesser extent from their parents. Genetics, as a science, makes it possible to predict and study the transmission of successive characters.
Instructions
1. Certain types of research are used to solve genetic problems. The method of hybridological review was developed by G. Mendel. It allows us to identify the validity of the inheritance of individual characters during sexual reproduction of organisms. The essence of this method is primitive: when reviewing certain alternative signs, their transmission in offspring is traced. Precise control of the manifestation of any alternative sign and the disposition of each individual offspring is also carried out.
2. The basic rationale for inheritance was also developed by Mendel. The scientist derived three laws. Later they were called that - Mendel's laws. 1st is the law of uniformity of first generation hybrids. Take two heterozygous individuals. When crossed they will produce two types of gametes. The offspring of such parents will appear in a ratio of 1:2:1.
3. Mendel's 2nd law is the law of splitting. it is based on the statement that a dominant gene does not invariably suppress a recessive one. In this case, not all individuals among the first generation reproduce the signs of their parents - the so-called intermediate inheritance pattern arises. For example, when crossing homozygous plants with red flowers (AA) and white flowers (aa), offspring with pink ones are obtained. Incomplete dominance is a fairly common phenomenon. It is also found in some human biochemical signs.
4. The 3rd and final law is the law of independent combination of signs. For this law to manifest itself, several conditions must be met: there should be no lethal genes, dominance must be complete, genes must be located on different chromosomes.
5. The problems of gender genetics stand apart. There are two types of sex chromosomes: the X chromosome (female) and the Y chromosome (male). A sex that has two identical sex chromosomes is called homogametic. Sex determined by different chromosomes is called heterogametic. The sex of the future individual is determined at the moment of fertilization. Sex chromosomes, in addition to genes that carry information about sex, also contain others that have nothing to do with this. Let's say the gene responsible for blood clotting is carried by the female X chromosome. Gender-linked signs are passed on from mother to sons and daughters, but from father - only to daughters.
All tasks By biology are divided into tasks by molecular biology And tasks by genetics. In molecular biology there are several topics that have tasks: proteins, nucleic acids, DNA code and energy metabolism.
Instructions
1. Decide tasks on the topic “Proteins” with the support of the following formula: m(min) = a/b*100%, where m(min) is the minimum molecular weight of the protein, a is the nuclear or molecular weight of the component, b is the percentage table of contents of the component. The average molecular weight of one acid residue is 120.
2. Calculate the required quantities on the topic “Nucleic acids”, adhering to Chargaff’s rules: 1. The number of adenine is equal to the number of thymine, and guanine is equal to cytosine; 2. The number of purine bases is equal to the number of pyrimidine bases, i.e. A+G = T+C. In the chain of a DNA molecule, the distance between nucleotides is 0.34 nm. The relative molecular weight of one nucleotide is 345.
3. Solve problems on the topic “DNA Code” with the help of a special table of genetic codes. As a result, you will find out which acid is encoded by one or another genetic code.
4. Calculate the result you need for problems on the topic “Energy Exchange” using the reaction equation. One of the frequently encountered ones is: C6H12O6 + 6O2? 6CO2 + 6H2O.
5. Find solutions to genetics problems using a special algorithm. First, determine which genes are dominant (A, B) and which are recessive (a, b). A gene is called dominant, the sign of which is manifested both in the homozygous (AA, aa) and heterozygous state (Aa, Bb). A gene is called recessive, the sign of which appears only when identical genes meet, i.e. in a homozygous state. Let's say pea plants with yellow seeds are crossed with pea plants with green seeds. The resulting pea plants all had yellow seeds. Apparently, the yellow color of the seeds is the dominant sign. Write down the solution to this tasks so: A is the gene responsible for the yellow color of seeds, and is the gene responsible for the green color of seeds. P: AA x aaG: A, aF1: AaExist tasks of this type with several signs, then designate one sign as A or a, and the 2nd sign as B or b.
Note!
Problems are a part of every biology course and it is very important to be able to solve them.
When comprehending genetics, great attention is paid to problems whose solutions must be discovered using the laws of gene inheritance. Most students of natural sciences solve problems in genetics seems to be one of the most difficult prophecies in biology. However, it is found using a simple algorithm.
You will need
- - textbook.
Instructions
1. To begin, read the problem carefully and write down a short schematic condition using special symbols. Indicate which genotypes the parents have and what phenotype corresponds to them. Write down what kind of children came out in the first and second generations.
2. Note which gene is dominant and which is recessive, if it is in the condition. If a split is given in the problem, also indicate it in the schematic notation. For primitive problems in genetics, it is sometimes quite positive to write down the condition in order to understand the solution tasks .
3. Start solving. Write down again the crossing that occurred: the genotypes of the parents, the resulting gametes and the genotypes (or supposed genotypes) of the children.
4. To successfully solve the problem, you need to understand which section it belongs to: monohybrid, dihybrid or polyhybrid crossing, sex-linked inheritance, or the sign is inherited through the interaction of genes. To do this, calculate what splitting of the genotype or phenotype is observed in the offspring in the first generation. The condition may indicate the exact number of individuals with each genotype or phenotype, or the percentage of the entire genotype (phenotype) of the total number. These data must be reduced to simple numbers.
5. Note whether the offspring have different signs depending on gender.
6. All types of crossing have their own special cleavage according to genotype and phenotype. All this data is contained in the textbook, and you will be comfortable writing these formulas on a separate sheet and applying them when solving problems.
7. Now that you have found the split, the thesis of which is the transfer of successive characters in your problem, you can find out the genotypes and phenotypes of all individuals in the offspring, as well as the genotypes and phenotypes of the parents who participated in the crossing.
8. Write the received data into the result.
Comprehension of genetics is accompanied by problem solving. They clearly show the effect of the law of genetic inheritance. Most students find solving these problems incredibly difficult. But, knowing the solution algorithm, you can easily cope with them.
Instructions
1. It is possible to distinguish two main types of genetic problems. In the first type of task, the genotypes of the parents are known. The genotypes of the offspring need to be determined. First determine which allele is dominant. Find the recessive allele. Record the genotypes of the parents. Write down all acceptable types of gametes. Combine the gametes. Define splitting.
2. In problems of the second type, everything is the opposite. Splitting in the offspring is famous here. It is necessary to determine the genotypes of the parents. Find out, in the same way as in problems of the first type, which of the alleles is dominant and which is recessive. Determine the acceptable types of gametes. Using them, determine the genotypes of the parents.
3. In order to solve the problem correctly, read it carefully and analyze the condition. To determine the type of problem, find out how many pairs of characters are considered in the problem. Pay attention also to how many pairs of genes control the formation of signs. It is important to find out whether homozygous or heterozygous organisms are crossed, what type of crossing is. Determine whether genes are inherited independently or linked, how many genotypes are formed in the offspring, and whether inheritance is related to gender.
4. Start solving the problem. Make a short note of the data. Record the genotype or phenotype of the individuals involved in the cross. Identify and note the types of gametes produced. Record the genotypes or phenotypes of the offspring obtained from the cross. Analyze the results, write them down in numerical terms. Write the result.
5. Remember that all types of crossing correspond to a special splitting according to genotype and phenotype. All this data can be found in textbooks or other aids. Write down all the formulas on a separate sheet of paper and keep it always at hand. You can also use special tables to solve genetics problems.
A tastefully decorated school office increases students' interest in the subject being studied and creates a comfortable learning atmosphere. How to design a biology classroom in such a way that students would be happy to come to lessons and it would be convenient for the teacher to teach in this room?
Instructions
1. Set up themed stands on one of the side walls of your office. They may bear the names: “Evolution of the Earth”, “Structure of the Cell”, “Let’s save for posterity, etc. Some stands may be equipped with replaceable panels. Make a stand with student work: reports, reports, drawings, fascinating photographs of nature, etc. Use a portable display with additional writing on the subject to increase student interest in biology.
2. Use portraits of famous scientists in the design of your biology classroom: C. Darwin, K. A. Timiryazev, I. M. Sechenov, I. P. Pavlov and others.
3. Install a TV with a DVD player and educational discs in the office so that students can watch educational thematic films.
4. Decorate your biology classroom with living vegetation: these can be indoor flowers, citrus fruits and palm trees grown in tubs. It is better to choose plants for classroom decoration that can be used as demonstration material in lessons.
5. Make a living corner in your biology classroom: place an aquarium with fish in it, house hamsters or a turtle; You can limit yourself to one aquarium.
6. Store visual aids used as teaching aids in biology lessons (collections, herbariums, models, etc.) on cabinet shelves or on separate hanging shelves. Tables and especially fragile and expensive materials (devices, models, raw preparations, etc.) are best stored in closed cabinets.
7. Choose a general color scheme for the cabinet in accordance with natural colors. For example, walls and curtains can be made in different shades of green, panels can be painted beige, the color that will echo the color of the furniture.
8. Draw a natural ornament with huge multi-colored circles (for example, flowers) on the ceiling of the office, one that can be used during physical education for eye training.
Helpful advice
Encourage students who are burdened to decorate the biology classroom with their crafts, drawings, etc., develop in them a sense of taste and love for nature.
G. Mendel used the hybridological method in his genetic skills. He crossed pea plants that differed in one or more characters. After this, the scientist analyzed the nature of the manifestation of signs in the offspring.
Instructions
1. Pure lines are varieties of plants that have a certain continuous sign, say, yellow or green seed color. Monohybrid crossing – crossing 2 pure lines of plants, differing only in one sign. In dihybrid crossing, individuals are taken from which differences in two signs are examined.
2. Let's say there is a clean line of peas with smooth yellow seeds, and a line with green and wrinkled ones. Signs are determined by pairs of genes, with one pair of genes encoding the color of the seeds, the other their shape. Yellow color and smooth shape are dominant genes, green color and wrinkled seeds are recessive.
3. In the first generation, all pea seeds will be yellow and smooth, according to the law of uniformity of first-generation hybrids. Here the phenomenon of complete dominance is monitored: only dominant genes are manifested, and recessive genes are suppressed.
4. For later solution tasks to dihybrid crossing you need to fill the Punnett grid. Plants of the first generation F1, merging with each other, will give four types of gametes: AB, Ab, aB and ab. Draw the frame of a four-by-four rectangular table. Label the gametes at the top of the columns. Write the gametes to the left of the lines in the same way. It's like playing naval battle.
5. All valid combinations of these four types of gametes will give in the second generation 9 different genotypes: AABB, AaBB, AABb, AaBb, aaBB, AAbb, aaBb, Aabb, aabb. But only four phenotypes will be tracked: yellow - smooth, yellow - wrinkled, green - smooth, green - wrinkled. The ratio of monitored phenotypes is 9:3:3:1.
6. If you look at the proportions between yellow and green peas separately, they will be 3:1, as is the case with monohybrid crossing m. The same applies to the smoothness or wrinkling of seeds.
7. It turns out that the splitting rule is carried out identically for mono- and dihybrid crossings. Therefore, it can be concluded that genes and the characters encoded by them during dihybrid crossing are inherited independently from each other. The law of independent inheritance of characters is objective only when genes are located on different non-homologous chromosomes.
Note!
Complete dominance is not consistently monitored. There are several other types: incomplete dominance, co-dominance, over-dominance.
Genetic information is the program of properties of a living organism, embedded in the form of a genetic code in successive structures (DNA molecules). The guess that similar information is recorded on the molecular level was first formulated by the outstanding Russian biologist N.K. Koltsov back in the 20s of the 20th century.
Instructions
1. Approximately every genetic information is located in the nuclei of cells of a living organism; DNA (deoxyribonucleic acid) macromolecules included in the chromosome structures are responsible for it. There is more than 2 meters of DNA in the human body. Using a special genetic code, they contain encrypted information about the structure of proteins, the structure and function of cells, the signs of each cell and, in total, each organism.
2. DNA programs the design of organic compounds, controls synthesis and breakdown. This information is transmitted during DNA replication (doubling). The system for recording it in a molecule is represented by a genetic code - information about the structure of the polypeptide molecule (types of amino acids, number and sequence of their arrangement). One gene contains data about one polypeptide chain. DNA molecules are essentially templates for the synthesis of proteins, all of which contain a lot of different genes that a person receives from his parents.
3. Scientists have found that all cells of one organism contain an identical set of genes. The performance of different functions by different cells is due to the fact that not every successive information is implemented in them, but only the necessary areas - genes. A similar process is performed during the synthesis of protein molecules using 3 RNAs: ribosomal (rRNA), messenger (mRNA), transport (tRNA). Information is transmitted through a direct communication channel (DNA - RNA - protein) and through a feedback channel (external environment - protein - DNA).
4. Living organisms can receive, store and transmit information, using it as efficiently as possible. Data embedded in genes is transmitted from an individual to his descendants. This information is used for formation and reproduction and determines the direction of formation of the organism. In the process of interaction of an individual with the environment, the reaction to it can be distorted, which ultimately ensures the evolution of the formation of descendants. At the same time, the cells “remember” new information.
5. When implementing a succession program, the phenotype of an organism of a given biological species is formed under certain environmental conditions, the morphological structure, formation, growth, metabolism, mental makeup, predisposition to disease, and successive defects of the body are determined. Many scientists emphasized the essential role of similar information in evolution and noted this circumstance as one of the main criteria of life.
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