The study of the basic laws of heredity and the 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 figure out how to solve problems in genetics.

The relevance of studying the patterns of heredity

The two most important branches of modern science - medicine and breeding - are developing thanks to the research of genetic scientists. The very same biological discipline, 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 color, blood type) will first have to study the laws of heredity and variability, and also figure out how to solve problems in human genetics. It is this question that we will deal with.

Basic concepts and terms

Each industry has a specific, unique to it, set of basic definitions. If we are talking about science that studies the processes of transmission of hereditary traits, we will understand the following terms as 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 problems in biology for genetics. But first we will study the hybridological method. After all, it is at the heart of genetic research. It was proposed by the Czech naturalist G. Mendel in the 19th century.

How are traits inherited?

The patterns of transferring the properties of an organism were discovered by Mendel thanks to the experiments that he conducted with a well-known plant - The hybridological method is a crossing of two units that differ from each other in one pair of characters (monohybrid crossing). If organisms that have several pairs of alternative (opposite) traits participate in the experiment, then they speak of polyhybrid crossing. The scientist proposed the following form of recording the course of hybridization of two pea plants, which differ in the color of the seeds. A - yellow paint, a - green.

In this entry F1 - hybrids of the first (I) generation. 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 one (Rule of Uniformity for F1 hybrids). Knowing it explains to students how to solve problems in genetics. Grade 9 has a program in biology, in which the hybridological method of genetic research is studied in detail. It also considers the second (II) Mendel's rule, called the law of splitting. According to him, in F2 hybrids obtained from crossing two first-generation hybrids with each other, splitting is observed in a ratio of 3 to 1 in phenotype, and 1 to 2 and 1 in genotype.

Using the above formulas, you will understand how to solve problems in genetics without errors, if under their conditions you can apply the first or already known II Mendel's law, given 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 seed shape, in plants such as garden peas, then the Pinnett grid should be used in the genetic cross.

Absolutely all hybrids that are the first generation obey Mendel's uniformity rule. That is, they are yellow, with a smooth surface. By continuing to cross between plants from F1, we will get second-generation hybrids. To figure out how to solve problems in genetics, grade 10 in biology lessons uses a dihybrid crossing record, using the phenotype splitting formula 9: 3: 3: 1. Provided that the genes are located in different pairs, you can use Mendel's third postulate - the law of independent combinations of trait states.

How are blood types inherited?

The mechanism of transmission of such a trait as a blood group in humans does not correspond to the patterns we have considered earlier. That is, it does not obey the first and second laws of Mendel. This is due to the fact that such a trait as a blood type, according to Landsteiner's research, is controlled by three alleles of the I gene: 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.
  • Fourth - AB.

Gene 0 is a recessive allele to genes A and B. And the fourth group is the result of codominance (the 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 for blood groups. But that's not all. To establish the genotypes of children by blood type 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 Trait States", in which we considered how to solve problems in genetics. like the third law of Mendel itself, to which it obeys, is applicable for allelic genes located in the homologous chromosomes of each pair.

In the middle of the 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 linkage groups. And their number is exactly equal to the haploid set of chromosomes. In the process of meiosis, leading to the formation of gametes, not individual genes enter the germ cells, as Mendel believed, but their entire complexes, called by Morgan linkage groups.

Crossing over

During prophase I (it is 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 the branches of biology dealing with the study of human hereditary diseases. Applying the postulates set out in Morgan's chromosome theory of heredity, we will define an algorithm that answers the question of how to solve problems in genetics.

Sex-linked cases of inheritance are a special case of the transfer of genes that are located on the same chromosome. The distance that exists between genes in linkage groups is expressed as a percentage - morganides. 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 recall 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 men, 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. Such hereditary human diseases as hemophilia and color blindness arise as a result of the “breakage” of individual genes on the X chromosome. For example, from the marriage of a carrier of hemophilia with a healthy man, the birth of such offspring is possible.

The above course of genetic crossing confirms the linkage of the gene that controls blood clotting with the sex X chromosome. This scientific information is used to teach students techniques that determine how to solve problems in genetics. Grade 11 has a biology program, which deals in detail with such sections as "genetics", "medicine" and "human genetics". They allow students to study human hereditary diseases and know the reasons why they occur.

Gene Interaction

The transmission of hereditary traits is a rather complicated process. The schemes given earlier become understandable only if students have a basic minimum of knowledge. It is necessary because it provides the mechanisms that answer the question of how to learn how to solve problems in biology. Genetics studies the forms of interaction of genes. These are polymeria, epistasis, complementarity. Let's talk about them in more detail.

An 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 - for the functioning of the auditory nerve. In the marriage of deaf parents, each of whom is a recessive homozygous for each one of the two pairs of genes, children with normal hearing are born. In their genotype, both dominant genes are present that control the normal development of the auditory apparatus.

Pleiotropy

This is an interesting case of gene interaction, in which the phenotypic manifestation of several traits at once depends on one gene present in the genotype. For example, in the west of Pakistan, human populations of some representatives have been found. 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 be formed in the process of ontogenesis.

Animals, such as Karakul sheep, have a dominant W gene that controls both fur color and normal stomach development. 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 anomalies in the development of the stomach. At the same time, ½ (having gray fur) are heterozygous and viable, and ¼ are individuals with black fur and normal stomach development (their WW genotype).

Genotype - an integral system

The multiple action of genes, polyhybrid crossing, the phenomenon of linked inheritance serve as indisputable proof of the fact that the totality of the genes of our body is an integral system, although it is represented by individual gene alleles. They can be inherited according to the laws of Mendel, independently or by loci, linked, 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 one or more traits that affect the development.

Task 12
In snapdragons, the red color of the flower incompletely dominates the white. The hybrid plant has a pink color. Narrow leaves incompletely dominate broad ones. In hybrids, the leaves are of medium width. What offspring will result from crossing a plant with red flowers and medium leaves with a plant having pink flowers and medium leaves?
Solution:
A - red color of the flower,
a - white flower color,
Aa - pink A flower color,
B - narrow leaves,
b - wide leaves,
Bb is the average width of the leaves.
The first plant with a red flower color is homozygous for the dominant trait, because with incomplete dominance, the plant with the dominant phenotype is heterozygous (AA). With incomplete dominance, the middle leaves have a plant that is heterozygous for the shape of the leaves (Bb), which means the genotype of the first plant is AABb (gametes AB, Ab).
The second plant is a diheterozygote, since it has an intermediate phenotype for both traits, which means its genotype is AaBb (gametes AB, Ab, aB, ab).

Crossing scheme

Answer:
25% - red flowers and medium leaves,
25% - pink flowers and medium leaves,
12.5% ​​- red flowers and narrow leaves,
12.5% ​​- pink flowers and narrow leaves,
12.5% ​​- pink flowers and broad leaves,
12.5% ​​- red flowers and broad leaves.

Task 13
It is known that the absence of stripes in watermelons is a recessive trait. What offspring will result from crossing two heterozygous plants with striped watermelons?
Solution:
A - watermelon banding gene
a - gene for lack of banding in watermelon
The genotype of a heterozygous plant is Aa (gametes A, a). when two heterozygotes are crossed, the offspring will show phenotypic splitting in a ratio of 3:1.
Crossing analysis confirms this statement.

Crossing scheme

Answer:
25% - plants with striped fruits with AA genotype,
50% - plants with striped fruits with Aa genotype,
25% - plants with stripless watermelons with aa genotype.

Task 14
In humans, the gene that causes one of the forms of hereditary deaf-mutism is recessive with respect to the gene for normal hearing. From the marriage of a deaf-mute woman with an absolutely healthy man, a healthy child was born. Determine the genotypes of all family members.
Solution:
A - gene for normal hearing development;
a - deaf-mute gene.
Since a woman suffers from deaf-mutism, her genotype is aa (gametes a). The man is absolutely healthy, which means he is homozygous for the dominant gene A, genotype AA (gametes A). In homozygous parents for the dominant and recessive gene (A), all offspring will be healthy.
Crossing analysis confirms this statement.

Crossing scheme

Answer:

1) genotype of a deaf-mute mother aa (gametes a),
2) father's genotype AA (gametes A),
3) the genotype of the child is Aa.

Task 15
Polled (hornless) in cattle dominates over horniness. The polled bull was crossed with a horned cow. From crossing, two calves appeared - horned and polled. Determine the genotypes of all animals.
Solution:
A - gene for polled (hornless) cattle;
a - hornedness gene.
This task is for monohybrid crossing, since the crossed organisms are analyzed for one pair of traits.
Since offspring appeared from crossing a horned bull and a horned cow - a horned and horned calf, the horned bull was heterozygous for the gene (A), because in a horned calf one gene (a) appeared from a horned cow, and the other from a horned bull, which means genotypes of parents: polled bull - Aa (gametes A, a), cow - aa (gametes a). From crossing a heterozygous bull with a cow homozygous for the recessive gene, offspring can appear according to the phenotype in a ratio of 1: 1.
Crossing analysis confirms this statement.

Crossing scheme

Answer:
The scheme for solving the problem includes:
1) cow genotype aa (gametes a),
2) bull genotype Aa (gametes A, a),
3) genotype of the comologist calf Aa,
4) the genotype of the horned calf aa.

Task 16
It is known that one of the forms of schizophrenia is inherited as a recessive trait. Determine the probability of having a child with schizophrenia from healthy parents, if it is known that they are both heterozygous for this trait.
Solution:
A - gene for normal development,
a - schizophrenia gene.
With monohybrid crossing of heterozygotes in the offspring, splitting is observed according to the genotype: 1:2:1, and according to the phenotype 3:1.
Crossing analysis confirms this statement.

Crossing scheme

Answer:
The probability of having a child with schizophrenia is 25%.

Problem 17
When gray flies were crossed with each other, splitting was observed in their F 1 offspring. 2784 were gray and 927 were black. What trait is dominant? Determine the genotypes of the parents.
Solution:
From the condition of the problem, it is easy to conclude that there are more gray individuals in the offspring than black ones, but because parents with gray coloration had cubs with black. On the basis of this, we introduce the conventions: gray color of flies - A, black - a.
There is a rule, if during a monohybrid crossing of two phenotypically identical individuals in their offspring, splitting of signs 3:1 (2784:927 \u003d 3:1) is observed, then these individuals are heterozygous.
Using the above rule, we can say that black flies (homozygous for a recessive trait) could only appear if their parents were heterozygous.
Let's check this assumption by constructing a crossover scheme:

Crossing scheme

Answer:
1) Gray color dominates.
2) Parents are heterozygous.

Problem 18
When crossing radish plants with oval roots, 66 plants with rounded, 141 with oval and 72 with long roots were obtained. How is the root shape inherited in radishes? What offspring will be obtained from crossing plants with oval and rounded roots?
Solution:
The ratio of offspring by phenotype in this cross is 1:2:1 (66:141:72 1:2:1). There is a rule: if when crossing phenotypically identical (one pair of traits) individuals in the first generation of hybrids, the traits are split into three phenotypic groups in a 1:2:1 ratio, then this indicates incomplete dominance and that parental individuals are heterozygous. According to this rule, in this case, the parents must be heterozygous.
Crossing analysis confirms this statement.

First cross scheme

Given that when plants with oval roots were crossed with each other, twice as many plants with oval roots appeared in the offspring, the genotype of plants with oval roots is Aa (gametes A, a), and the genotype of plants with rounded roots is AA (gametes A). We define the offspring that will result from crossing plants with oval and rounded roots.

Scheme of the second crossing

Answer:
1) Inheritance is carried out according to the type of incomplete dominance.
2) When crossing plants with oval and rounded roots, 50% of plants with oval and 50% with rounded roots will be obtained.

Problem 19
In humans, brown-eyedness dominates over blue-eyedness, and dark hair color over light. A blue-eyed, dark-haired father and a brown-eyed, blond mother have four children. Each child is different from the other in one of these ways. What are the genotypes of parents and children?
Solution:
A - gene for cross-eyedness,
a - blue-eyed gene,
B - dark hair
b - blond hair.
The mother is homozygous for the recessive trait of blond hair (bb), and the father is homozygous for the recessive trait of light eyes (aa). Since splitting is observed in the offspring for each trait, organisms that exhibit dominant traits are heterozygous for the genes encoding it. Then the genotypes of the parents: mother - Aabb (gametes Aa, ab), father - aaBb (gametes aB, ab).
Let's determine the genotypes of the offspring:

Crossing scheme

Answer:
1) For each of the traits, splitting occurs in the offspring, therefore, organisms showing a dominant trait are heterozygous for the genes encoding it. Therefore, the mother's genotype is Aaaa (gametes Aa, aa), and the father's genotype is aaBb (gametes aB, ab).
2) father and mother produce two types of gametes each, which give 4 variants of combinations. Hence. the genotype of children is aabb, aaBb, Aabb, AaBb.

Problem 20
In chickens, the black color of plumage dominates over red, the presence of a crest over its absence. The genes encoding these traits are located on different pairs of chromosomes. A red rooster with a comb is crossed with a black hen without a comb. Numerous offspring were obtained, half of which have black plumage and crest, and half - red plumage and crest. What are the genotypes of the parents?
Solution:
A - black plumage gene,
a - red plumage gene
B - the gene responsible for the formation of the ridge
b - gene responsible for the absence of a crest.
The rooster is homozygous for the recessive gene for plumage color (aa), and the hen is homozygous for the recessive gene for ridge formation (bb). Since, according to the dominant trait of plumage color (A), half of the offspring are black, half are red, the black chicken is heterozygous for plumage color (Aa), which means its genotype is Aabb. According to the dominant trait of the formation of the crest, all offspring have a comb, which means that the rooster is homozygous for the presence of the BB crest). Therefore, the rooster genotype is aaBB.
The analysis of the conducted crossing confirms our reasoning.

Crossing scheme

Answer:
1) The rooster genotype is aaBB.
2) Chicken genotype Aabb.

Problem 21
Two breeds of silkworms were crossed, which differed in two ways: striped caterpillars wove white cocoons, and monochromatic caterpillars wove yellow cocoons. In the F 1 generation, all caterpillars were striped and weaving yellow cocoons. In the F 2 generation, splitting was observed:
3117 - striped caterpillars weaving yellow cocoons,
1067 - striped caterpillars spinning white cocoons,
1049 - single color with yellow cocoons,
351 - single color with white cocoons.
Determine the genotypes of the original forms and offspring F 1 and F 2 .
Solution:
This task is for dihybrid crossing (independent inheritance of traits during dihybrid crossing), since caterpillars are analyzed according to two traits: body color (striped and one-color) and cocoon color (yellow and white). These traits are due to two different genes. Therefore, to designate genes, we take two letters of the alphabet: “A” and “B”. Genes are located in autosomes, so we will designate them only with the help of these letters, without using the symbols X- and Y- chromosomes. The genes responsible for the analyzed traits are not linked to each other, so we will use the genetic record of the crossing. either dominant or recessive to each other. First, we determine which traits are dominant and which are recessive. In the F 1 generation, all silkworm caterpillars were striped and weaving yellow cocoons, which means that the stripedness of the caterpillars (A) is a dominant trait, and the uniform color (a) is recessive, and the yellow color (B) dominates over white (b). From here:
A - caterpillar striping gene;
a - gene for the uniform color of caterpillars;
B - yellow cocoon gene;
b - white cocoon gene.
Let's determine the genotypes of the offspring:

First cross scheme

biology, exam, recessive trait, polledness, schizophrenia, dihybrid crossing, autosomal dominant type of inheritance of a trait, autosomal recessive type of inheritance of a trait, proband.
The genotype of the F 1 offspring is AaBb (gametes AB, Ab, aB, ab).
According to Mendel's third law, during dihybrid crossing, the inheritance of both traits is carried out independently of each other, and in the offspring of diheterozygotes, phenotypic splitting is observed in the proportion 9:3:3:1 (9 A_B_, 3 aaB_, 3 A_bb, 1 aabb, where ( _ ) in this case means that the gene can be either in a dominant or recessive state). According to the genotype, splitting will be carried out in a ratio of 4:2:2:2:2:1:1:1:1 (4 AaBb, 2 AABb, 2 AaBB, 2 Aabb, 2 aaBb, 1 AAbb, 1 AABB, 1 aaBB, 1 aabb).
Cross-breeding analysis confirms these considerations.
Now let's determine the genotypes of the offspring by analyzing the crossing of parent plants:

Scheme of the second crossing

Answer:
1) The genes for the striped color of the caterpillars and the yellow color of the cocoons are dominant. According to Mendel's first law, the genotypes of the original forms (P) are AAbb (Ab gametes) and aaBB (aB gametes), uniform offspring F 1 - AaBb (gametes AB, Ab, aB, ab).
2) In the offspring of F 2, a splitting close to 9:3:3:1 is observed. Striped individuals with yellow cocoons had genotypes 1AABB, 2AaBB, 2AABb, 4AaBb. Striped with white cocoons AAbb, 2Aabb, single-colored with yellow cocoons - aaBB and 2aaBb, single-colored with white cocoons aabb.

Problem 22
According to the pedigree shown in the figure (Fig. 1.), 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.

Rice. 1. Graphical representation of a pedigree for an autosomal dominant type of inheritance of a trait, consisting of three generations

Solution:





- marriage of a man and a woman;
- consanguineous marriage;

- childless marriage;

People with the trait under study are found frequently, in every generation; a person with a studied trait is born in a family where at least one of the parents must have the studied trait. Therefore, we can make the first preliminary conclusion: the trait under study is dominant. In the pedigree, 2 women and 2 men have the studied trait. It can be assumed that the trait under study occurs with approximately equal frequency in both men and women. This is typical for traits whose genes are located not on the sex chromosomes, but on the autosomes. Therefore, we can make a second preliminary conclusion: the trait under study is autosomal.
Thus, according to the main features, the inheritance of the studied trait in this pedigree can be attributed to the autosomal dominant type. In addition, this pedigree does not have a set of features that are characteristic of other types of inheritance.

According to the pedigree scheme, the man is sick, and the woman is healthy, they had three children - one is healthy, and two are sick, this suggests that all individuals with the studied trait are heterozygous. Then the genotypes of the members of the pedigree:
children of the 1st generation: daughter of Aa, daughter of aa, son of Aa;
2nd generation children: Aa's daughter;
mother aa, father aa.
Answer:
1) the trait is dominant, not sex-linked;
2) the genotype of children of the 1st generation: daughter Aa, daughter aa, son Aa;
3) genotype of children of the 2nd generation: daughter of Aa.

Problem 23
According to the pedigree shown in the figure, set the nature of the manifestation of the trait (dominant, recessive), indicated in black. Determine the genotype of parents and children in the first and second generation.

Rice. 2. Graphical representation of a pedigree for an autosomal recessive type of inheritance of a trait, consisting of three generations

Solution:
Symbols used in drawing up a graphic representation of a pedigree:
- a male individual that does not have the trait under study;
- a female individual that does not have the trait under study;
- a male individual with the trait under study;
- a female individual with the trait under study;
- marriage of a man and a woman;
- consanguineous marriage;
- children of one parental couple (siblings);
- childless marriage;

People with the studied trait are rare, not in every generation. Therefore, we can make the first preliminary conclusion: the trait under study is recessive. In the pedigree, 1 female and 1 male have the studied trait. It can be assumed that the trait under study occurs with approximately equal frequency in both men and women. This is typical for traits whose genes are located not on the sex chromosomes, but on the autosomes. Therefore, we can make a second preliminary conclusion: the trait under study is autosomal.
Thus, according to the main features, the inheritance of the studied trait in this pedigree can be attributed to the autosomal recessive type. In addition, this pedigree does not have a set of features that are characteristic of other types of inheritance.
Let's determine the possible genotypes of all members of the pedigree:
According to the pedigree scheme, the man is healthy, and the woman is sick, they had two children - the girl is healthy, and the boy is sick, this suggests that all individuals with a phenotype for the studied trait are homozygous (aa), and healthy family members are heterozygous (Aa) . Then the genotypes of the members of the pedigree:
children of the 1st generation: daughter of Aa, son of aa;
3) children of the 2nd generation: son of Aa, daughter of Aa;
mother aa, father Aa or AA.
Answer:
1) the trait is recessive, not sex-linked;
2) genotypes of parents: mother - aa, father - AA or Aa;
3) genotype of children of the 1st generation: daughter Aa, son aa;
3) genotype of children of the 2nd generation: daughter of Aa, son of Aa.

"FABLE" TASKS IN GENETICS

Although the tasks are fabulous, everything in them obeys known genetic laws. Solving such problems is more interesting than the usual ones about peas and a cockscomb.

Monohybrid cross

1. Flowertails live on asteroid 2244-P. Their coloration is yellow, green and green with yellow polka dots. When a green male was crossed with a yellow female, the cubs were hatched green, yellow and green with yellow polka dots. How is coloration inherited in flowertails if green is dominant? What are the genotypes of parents and offspring?

2. The Firebird has a bright yellow plumage, the Blue Bird is blue. When the Firebird was crossed with the Blue Bird, blue chicks hatched. What are the genotypes of parents and offspring? What is the dominant trait?

3. When crossing Scarlet Flowers with red and pink petals (there are also such), offspring with red and white flowers were obtained. Explain your results.

4. In Koloboks, the baldness gene dominates the hairiness gene. A hairy kolobikh, having a bald brother and a bald father, rolled out to marry a bald kolobok. They had a bald bun. The bun rolled out behind the hairy bun. What is the probability that they will have a bald bun; hairy bun?

5. Snow White has the second blood type (A), and the prince has the third (B). What is the probability of the birth of an heir with the first blood group (0), in which case is this possible?

6. The Little Humpbacked Horse was born to a mare of normal height. What are the genotypes of the parents if normal height is a dominant trait and the Little Humpbacked Horse had two brothers also of normal height?

7. In Jack and the Beanstalk, Jack planted a bean that grew into a sky-high plant. It turned out that the bean seed was obtained by crossing plants of normal growth. Such giant seeds were only a part of all received. What are the genotypes of the parental forms of plants?

8. The gene that determines laziness dominates performance. There is a suspicion that Emelya from the fairy tale "By the Pike's Command" is heterozygous. What is the percentage of this probability, if it is known that Emelya's mother was a hard worker, and her father was very lazy?

9. On the planet Vester, the inhabitants are very similar to people, but differ only in that they have lilac eyes and six fingers on their hands. A woman with purple eyes and six fingers married an Earthman with blue eyes and five fingers. They had a daughter with lilac eyes and five fingers. What is the probability that the other child will also be the same? What are the genotypes of parents and children?

10. In unicorns from the planet Krina, the white color depends on the dominant gene B, and yellow on its recessive allele b. Trotting is dependent on the dominant P gene, while walking is dependent on the recessive p gene. What will be the first generation phenotype when a homozygous white ambling unicorn is crossed with a homozygous yellow trotter? What offspring and in what proportions will be obtained by crossing two individuals of the first generation?

11. The inhabitants of the planet Bluk have big ears, so they are called Ushans. Ushanka with three ears marries Ushan with three ears, one of whose parents had two ears. From this marriage, one child was born with two ears. Make a family tree. Determine the genotypes.

12. Chernomor has a small stature, his brother was a giant, and his parents and sister were of normal human height. What are the genotypes of all members of the family, if high growth is a dominant trait?

13. The kid from the fairy tale "The Kid and Carlson" was given a white puppy with long hair for his birthday. The puppy's parents were both black with short hair. They had two puppies: black with short hair and white with long hair. Explain which traits are recessive and which are dominant. How can this be checked? What are the genotypes of all members of this family?

14. A miller died in the village.

Burying my father

The inheritance was divided

Three young brothers.

The elder brother took the mill,

The second one took the donkey,

And the cat went to the youngest,

The cat was adopted by the younger brother.

Did the brothers divide the inheritance according to the law, because many neighbors believed that not all the brothers were the miller's own sons? Is it possible, on the basis of blood groups, to consider the brothers the sons of a miller? The blood types are as follows: miller - 0A, mother - AB, first son - 00, second son - AA, third son - 0B.

15. Thumb-Boy has AB blood type, his mother has 0A blood type. What blood type can a father have?

Inheritance of sex-linked traits

1. In birds, the female sex is heterogametic. In Ryaba Hen, the sex-linked gene D determines the black color, and the d gene determines the pockmarked. They crossed a black hen with a pockmarked rooster. What are the genotypes of parents and chicks? What is the color of these chicks? How many of them will be speckled chickens?

2. Cheburashka's big ears are determined by recessive genes located in

X chromosome. Cheburashka with big ears marries Cheburakh with small ears

mi. What is the probability of appearance in the family of Cheburashkas and Cheburakhs with big ears?

3. One of Kir Bulychev's stories tells about the pirate Rat. This pirate could turn into different people. His dad could do it too, but his mom couldn't. It turned out that in the family all men were able to do this, since the transformation gene is sex-linked. Explain why only the males in the family had the ability to transform.

4. Carlson loves sweets very much. The gene that determines this trait is recessive and is located on the X chromosome. Carlson's grandmother also has a sweet tooth, but her mother does not. Write down the genotypes of the family. If Carlson had siblings, what is the probability that they would have a sweet tooth? Solve two versions of the problem: dad likes sweets and does not like.

5. Ivanushka caught the Firebird when he poured grains with wine into the trough. One of the birds got caught. It turns out that the wine love gene is recessive and is located on the X chromosome. Determine the genotype and sex of the Firebird, if the female gender, like all birds, is heterogametic.

6. Nightingale the robber and his father have a very powerful voice. The mother has a normal voice. Would it be correct to say that in this family the son inherited his father's voice if the nightingale whistle gene is on the X chromosome?

7. The gene for the gold content of eggs is located on the X chromosome. Hen Ryaba can lay golden and simple eggs. Moreover, golden eggs are obtained only if both X chromosomes carry this gene. What should be the genotypes of the Hen and the Rooster to make the egg golden?

8. In Goldfish, the a gene is recessive, sex-linked and lethal. What will be the numerical ratio of the sexes in the offspring from crossing the female Aa with a normal male?

9. When crossing a red-eyed female Goldfish with a blue-eyed male, 8 males with red eyes and 8 females with blue eyes were obtained. Determine the genotypes.

Interaction of non-allelic genes

1. On the planet Brasak (from the story of Kira Bulychev), the inhabitants of Brasak had blue and yellow eyes. Moreover, it turned out that the blue-eyed brasaki were honest, and those with yellow eyes were traitors who helped the pirates. Eye color is inherited by two pairs of genes, and the eyes are blue, yellow, orange. The brighter the tendency to betrayal was manifested, the more yellow were the eyes. Explain how eye color is inherited in Brasacs.

2. On the planet of the Blue Sun, very beautiful flowers grow, having a different shape of the corolla. When crossing plants with a funnel-shaped corolla (A) with a plant having a goblet shape (B), in the first generation all hybrids were cup-shaped. The recessive allele determines the cylindrical shape. Determine how the corolla shape is inherited in plants and what hybrids will be obtained in the second generation when crossing hybrids of the first generation.

3. On the third planet of the Medusa system, Alisa Selezneva discovered flowers with a mirror core, which is inherited by two pairs of genes. It turned out that mirror cores are obtained by crossing two plants with ordinary cores. How is specularity inherited in a plant? What are the genotypes of the plant?

4. Birds Talkers are smart and quick-witted. The tufts on the head are white and red, the color is inherited by two pairs of genes. Two white-tufted birds interbreed, and they hatch a chick with a red tuft on its head. How is the color of the tuft in Govoruns inherited? What are the genotypes of the parents and the chick?

5. Exploring the newly discovered planet, scientists discovered two races of the sweet bell plant, which have white flowers. Later, by crossing them in the first generation, all plants with purple flowers were obtained, and in the second they received splitting: 53 purple, 19 blue, 56 white. How can this fact be explained?

6. When crossing Tigerrats of black and red color from the planet Penelope in the first generation, approximately 1/2 red, 3/8 black and 1/8 brown were obtained. How is coat color inherited? What are the genotypes of parents and offspring?

7. When a dragon with three heads was crossed with a dragon with six heads, nine-headed dragons were obtained. When crossing them among themselves in the second generation, dragons of all four forms were found: 6 nine-headed, 4 six-headed, 3 three-headed and 1 twelve-headed. Explain your results.

1. The subject of genetics. Methods and tasks of genetics. Genetics is the theoretical basis of breeding and seed production.

2. A brief history of the development of genetics.

3. The structure of the cell. The role of cellular organelles in heredity.

4. Chromosomes, their morphology and biological significance. The concept of karyotype.

5. Transfer of hereditary information in the process of cell division by mitosis. phases of mitosis. The biological significance of mitosis.

6. Transfer of hereditary information during sexual reproduction. Reduction and equation division. The biological significance of meiosis.

7. Microsporogenesis and microgametogenesis in plants.

8. Macrosporogenesis and macrogametogenesis in plants.

9. Double fertilization in angiosperms. Apomixis, its types.

10. Method of genetic analysis developed by G. Mendel. genetic symbolism. Recording crosses and their results.

11. Monohybrid crossing, I and II laws of Mendel.

12. Analyzing, reciprocal and backcrosses, their significance for genetic breeding research.

13. Dihybrid and polyhybrid crosses. Third law of Mendel.

14. Chromosomal theory of heredity. Types of sex determination.

15. Inheritance of sex-linked traits.

16. Influence of factors of the internal and external environment on the development of sex characteristics. Experimental change in sex ratio.

17. Inheritance of traits with full linkage of genes.

18. Inheritance of traits with incomplete linkage of genes.

19. Crossing over, its types and factors affecting chromosome crossing.

20. Studies that have established the role of nucleic acids in heredity (transformation in bacteria, transduction).

21. Structure, functions of DNA, DNA replication.

22. Structure, functions and types of RNA in the cell, features of their structure.

23. Genetic code, its properties.

24. Matrix principle of heredity realization: transcription, translation.

25. Protein synthesis in a cell, its stages.

26. Regulation of protein synthesis.

27. The structure of the eukaryotic gene: exons, introns.

28. Genetic engineering. Methods for obtaining genes and their transfer.

29. Cytoplasmic male sterility (CMS).

30. Types of variability. modification variability.

31. Mutational variability. The main provisions of the mutation theory.

32. Natural (spontaneous) mutagenesis, its factors.

33. Classification of mutations according to their effect on the body.

34. Classification of mutations according to their effect on cell structures: genomic and chromosomal mutations.

35. Gene mutations. Molecular mechanism of gene mutations. Transitions, transversions.

36. Induced physical and chemical mutagenesis, its use in practical breeding.


37. Repair of damage to genetic material. Dark repair, repair enzymes.

38. Multiple allelism. The law of homological series of variability N.I. Vavilov. The use of artificial mutagenesis in plant breeding.

39. Autopolyploidy, features of meiosis and the nature of splitting in tetraploids. Triploidy, its use in practical breeding.

40. Allopolyploidy, its role in evolution and plant breeding.

41. Aneuploidy, haploidy. Their production and use in genetic breeding research.

42. Non-crossing of species, its causes. Methods for overcoming non-crossing.

43. Infertility of distant hybrids, its causes and ways of overcoming.

44. Inbreeding, inbreeding, its genetic essence. Characteristics of induction lines, their production and practical use.

45. Phenomenon of heterosis, types of heterosis, theories of heterosis.

46. ​​Practical use of heterosis on the example of obtaining double interline hybrids of corn using CMS.

47. The concept of populations. Population dynamics in self-pollinators.

48. Dynamics of populations at cross-breeds. Hardy-Weinberg law.

49. Changes in the genetic composition of populations under the influence of mutations, migrations and selection.

50. Changes in the genetic composition of populations under the influence of genetic drift and isolation.

Among the tasks in genetics at the exam in biology, 6 main types can be distinguished. The first two - to determine the number of types of gametes and monohybrid crossing - are most often found in part A of the exam (questions A7, A8 and A30).

Tasks 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 exam.

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

Basic terms of genetics

Gene- This is a section of the 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 the same gene encoding an alternative manifestation of the same trait. Alternative signs - signs that cannot be in the body at the same time.

Homozygous organism- an organism that does not give splitting for one reason or another. Its allelic genes equally affect the development of this trait.

heterozygous organism- an organism that gives splitting according to one or another feature. Its allelic genes affect the development of this trait in different ways.

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

recessive gene is responsible for the trait, the development of which is suppressed by the dominant gene. A recessive trait appears 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 uniformity of hybrids

This law is derived on the basis of the results of monohybrid crossing. For the experiments, two varieties of peas were taken, differing from each other in one pair of traits - the color of the seeds: one variety had a yellow color, the second - green. Crossed plants were homozygous.

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

Yellow seed color
- green seed color

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

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

Mendel's second law - the law of splitting

Plants were grown from seeds obtained by crossing a homozygous plant with yellow seeds with a plant with green seeds, and by self-pollination was obtained.


(plants have a dominant trait, - recessive)

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

Mendel's third law - the law of independent inheritance

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

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

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


(yellow smooth).

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

The Punnett grid is used to record and determine genotypes.
Gametes

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

When analyzing the inheritance of each pair of traits, the following results are obtained. In parts of yellow seeds and parts of green seeds, i.e. ratio . Exactly the same ratio will be for the second pair of characters (seed shape).

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

Mendel's third law holds only if the genes are on different pairs of homologous chromosomes.

Law (hypothesis) of "purity" of gametes

When analyzing the characteristics of hybrids of the first and second generations, Mendel found that the recessive gene does not disappear and does not mix with the dominant one. In both genes are manifested, which is possible only if the hybrids form two types of gametes: one carries a dominant gene, the other 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 proved after studying the processes occurring in meiosis.

The hypothesis of "purity" of gametes is the cytological basis of Mendel's first and second laws. With its help, splitting by phenotype and genotype can be explained.

Analyzing 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 it could be concluded that the original organism is homozygous for the trait under study.

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

Inheritance of blood groups (AB0 system)

The inheritance of blood groups in this system is an example of multiple allelism (the existence of more than two alleles of one gene in a species). There are three genes in the human population that code for erythrocyte antigen proteins that determine people's blood types. The genotype of each person contains only two genes that determine his blood type: the first group; second and ; third and fourth.

Inheritance of sex-linked traits

In most organisms, sex is determined at the time of fertilization and depends on the set of chromosomes. This method is called chromosomal sex determination. Organisms with this type of sex determination have autosomes and sex chromosomes - and.

In mammals (including humans), the female sex has a set of sex chromosomes, the male sex -. The female sex is called homogametic (forms one type of gametes); and male - heterogametic (forms two types of gametes). In birds and butterflies, males are homogametic and females are heterogametic.

The USE includes tasks only for traits linked to the -chromosome. Basically, they relate to two signs of a person: blood clotting (- normal; - hemophilia), color vision (- normal, - color blindness). Tasks for the inheritance of sex-linked traits in birds are much less common.

In humans, the female sex may be homozygous or heterozygous for these genes. Consider the possible genetic sets in a woman on the example of hemophilia (a similar picture is observed with color blindness): - healthy; - healthy, but is a carrier; - sick. The male sex for these genes is homozygous, tk. - chromosome does not have alleles of these genes: - healthy; - is sick. Therefore, men are most often affected by these diseases, and women are their carriers.

Typical USE tasks in genetics

Determination of the number of types of gametes

The number of gamete types is determined by the formula: , where is the number of gene pairs in the heterozygous state. For example, an organism with a genotype has no genes in a heterozygous state; , therefore, and it forms one type of gamete. 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.

Tasks for mono- and dihybrid crossing

For a monohybrid cross

Task: Crossed white rabbits with black rabbits (black color is a dominant trait). In white and black. Determine the genotypes of parents and offspring.

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

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

For a dihybrid cross

Dominant genes are known

Task: Crossed tomatoes of normal growth 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 the red color of the fruit dominates over yellow, and normal growth over dwarfism.

Solution: Denote dominant and recessive genes: - normal growth, - dwarfism; - red fruits, - yellow fruits.

Let us analyze the inheritance of each trait separately. All offspring have normal growth, i.e. splitting on this basis is not observed, so the original forms are homozygous. Splitting is observed in fruit color, so the original forms are heterozygous.
(dwarfs, red fruits)
(normal growth, red fruits)
(normal growth, red fruits)
(normal growth, red fruits)
(normal growth, yellow fruits)
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 red saucers, red funnels, white saucers and white funnels. Determine the dominant genes and genotypes of parental forms, as well as their descendants.

Solution: Let us analyze the splitting for each feature separately. Among the descendants, plants with red flowers are, with white flowers -, i.e. . Therefore, red - white color, and parental forms are heterozygous for this trait (because there is splitting in the offspring).

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


(red flowers, saucer-shaped)
(red flowers, funnel-shaped)
Gametes

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 groups are possible in children?

Solution:


(the probability of having a child with the second blood type is , with the third - , with the fourth - ).

Solving problems on the inheritance of sex-linked traits

Such tasks may well occur both in part A and in part C of the USE.

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 ()

Solving problems of mixed type

Task: A man with brown eyes and blood type married a woman with brown eyes and 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 may have the genotype or, the first - only. Since the child has the first blood group, therefore, he received the gene from both his father and mother, therefore his father has a genotype.

(father) (mother)
(was born)

Task: The 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, the best possession of the right hand dominates over left-handedness, therefore - right-handed, - lefty. Male genotype (because he received the gene from a left-handed mother), and women -.

A color-blind man has the genotype, and his wife -, 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 ()

Tasks for independent solution

  1. Determine the number of types of gametes in an organism with a genotype.
  2. Determine the number of types of gametes in an organism with a genotype.
  3. They crossed tall plants with short plants. B - all plants are medium in size. What will be?
  4. They crossed a white rabbit with a black rabbit. All rabbits are black. What will be?
  5. They crossed two rabbits with gray wool. B with black wool, - with gray and white. Determine the genotypes and explain this splitting.
  6. They crossed a black hornless bull with a white horned cow. They received black hornless, black horned, white horned and white hornless. Explain this split if black and the absence of horns are dominant traits.
  7. They crossed Drosophila with red eyes and normal wings with Drosophila 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?
  8. A blue-eyed brunette married a brown-eyed blonde. What kind of children can be born if both parents are heterozygous?
  9. A right-handed man with a positive Rh factor married a left-handed woman with a negative Rh factor. What kind of children can be born if a man is heterozygous only for the second trait?
  10. The mother and father have a blood type (both parents are heterozygous). What blood group is possible in children?
  11. The mother has a blood group, the child has a blood group. What blood type is impossible for a father?
  12. The father has the first blood type, the mother has the second. What is the probability of having a child with the first blood group?
  13. A blue-eyed woman with a blood type (her parents had a third blood type) married a brown-eyed man with a blood type (his father had blue eyes and a first blood type). What kind of children can be born?
  14. A right-handed hemophilic man (his mother was left-handed) married a left-handed woman with normal blood (her father and mother were healthy). What kind of children can be born from this marriage?
  15. Strawberry plants with red fruits and long-leaved leaves were crossed with strawberry plants with white fruits and short-leaved leaves. What offspring can there be if red color and short-leaved leaves dominate, while both parental plants are heterozygous?
  16. A man with brown eyes and blood type marries a woman with brown eyes and blood type. They had a blue-eyed child with a blood type. Determine the genotypes of all individuals indicated in the problem.
  17. They crossed melons with white oval fruits with plants that had white spherical fruits. The following plants were obtained in the offspring: with white oval, with white spherical, with yellow oval and with yellow spherical fruits. Determine the genotypes of the original plants and descendants, if the white color of the melon dominates over the yellow, the oval shape of the fruit is over the spherical.

Answers

  1. gamete type.
  2. gamete types.
  3. gamete type.
  4. high, medium and low (incomplete dominance).
  5. black and white.
  6. - black, - white, - grey. incomplete dominance.
  7. Bull:, cow -. Offspring: (black hornless), (black horned), (white horned), (white hornless).
  8. - Red eyes, - white eyes; - defective wings, - normal. Initial forms - and, offspring.
    Crossing results:
    a)
  9. - Brown eyes, - blue; - dark hair, - light. Father mother - .
    - brown eyes, dark hair
    - brown eyes, blonde hair
    - blue eyes, dark hair
    - blue eyes, blonde hair
  10. - right-handed, - left-handed; Rh positive, Rh negative. Father mother - . Children: (right-handed, Rh positive) and (right-handed, Rh negative).
  11. Father and mother - . In children, a third blood type (probability of birth -) or a first blood type (probability of birth -) is possible.
  12. Mother, child; He received the gene from his mother, and from his father -. The following blood types are impossible for the father: second, third, first, fourth.
  13. A child with the first blood group can only be born if his mother is heterozygous. In this case, the probability of birth is .
  14. - Brown eyes, - blue. Woman man . Children: (brown eyes, fourth group), (brown eyes, third group), (blue eyes, fourth group), (blue eyes, third group).
  15. - right-handed, - lefty. Man Woman . Children (healthy boy, right-handed), (healthy girl, carrier, right-handed), (healthy boy, left-handed), (healthy girl, carrier, left-hander).
  16. - red fruit - white; - short-stalked, - long-stalked.
    Parents: and Offspring: (red fruit, short stem), (red fruit, long stem), (white fruit, short stem), (white fruit, long stem).
    Strawberry plants with red fruits and long-leaved leaves were crossed with strawberry plants with white fruits and short-leaved leaves. What offspring can there be if red color and short-leaved leaves dominate, while both parental plants are heterozygous?
  17. - Brown eyes, - blue. Woman man . Child:
  18. - 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.