Types of tasks in genetics. Task C6. Solving problems in genetics using the laws of G. Mendel Task in genetics

with examples

explanations and solutions

to solve problems in genetics, it is necessary:

enter symbols for features. The dominant gene is indicated by a capital letter, and the recessive gene is indicated by a capital letter;

find out what crossing takes place in this case (mono-, di- or polyhybrid);

write down the solution of the problem schematically and draw a conclusion about the probability of the requested event in% or fractions of unity.

Monohybrid cross.

One pair of alternative (mutually exclusive) traits is taken into account, therefore, one pair of allelic genes.

Task #1

In tomatoes, the gene for the red color of the fruit dominates the gene for the yellow color. What color fruits will be in plants obtained by crossing heterozygous red-fruited plants with yellow-fruited ones? What are their genotypes?

Solving problem number 1

The gene responsible for the red color of the fruit is dominant, denoted by A, the gene for the yellow color of the fruit is recessive, denoted by - a. The task of monohybrid crossing, because in the condition, the plants differ in 1 pair of mutually exclusive characters (fruit color). One of the parent 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 gives 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 gives one type of gametes (a).

Knowing the genotypes of the parents, we write down the solution of the problem and answer the question posed.

A - red fruits

a - yellow fruits

R - parents

G - gametes

F 1 - first generation

x - cross sign

R Aa x aa

red yellow

Answer: F 1 hybrids show phenotypic splitting in a ratio of 1: 1 - 50% red-fruited, 50% yellow-fruited tomatoes. By genotype, splitting in a ratio of 1: 1 - 50% of heterozygous individuals (Aa), 50% of homozygous individuals (aa).

Tasks for independent solution.

3Task number 2.

In Drosophila, gray body color dominates over black. From the crossing of flies with a gray body and flies with a black body, hybrids F 1 were obtained, which subsequently, when crossed with each other, gave 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 among F 2 offspring?

3. How many different genotypes are there among F 2 offspring?

4. How many homozygous gray flies are in F 2 (theoretically)?

5. How many blacks are in F 2 (theoretically)?

Task number 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 formed 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 number 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 hypertension with early onset, 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 early hypertension.

1. How many different genotypes can there be among the children of the above spouses?

2. How many types of gametes does a daughter have?

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 grandchildren from a daughter?

The general biology course at the school includes sections on genetics and molecular biology. Both sections are difficult enough for students to understand. And in order to improve the assimilation of this material, they resort to the help of solving problems. Such a solution of problems in biology, genetics involves the use of methodological techniques, namely: it is necessary to carefully study the condition of the problem, determine its type and write down the crossing scheme.

Solving problems in genetics

Consider how to solve problems in genetics. While solving problems, students develop logical thinking, they begin to understand the material on the topic being studied more deeply. There are generally accepted notations that are used in solving problems:

The letter R. It is used to denote parents;
Descendants of the first generation - F1, the second generation - F2, etc.;
Crossing - x;
To designate a gene that exhibit dominant traits, use the letters - A, B, C, etc.;
The letters a, b, c, etc. are used to designate a gene that exhibits recessive traits.

In crossbreeding schemes, the mother's genotype is written on the left side, and the father's genotype is written on the right. If you solve problems in genetics, a monohybrid crossing in the record will look like this: AA x aa.

Dihybrid cross

All living organisms can differ from each other according to various traits and genes. In order to make an analysis of the inheritance of several traits, each pair requires studying its inheritance separately, and then comparing and combining all observations. If the parental forms have two pairs of different traits, then such a cross is called a dihybrid.

For example, tasks on genetics of dihybrid crossing will look like this: in a family where the mother and father are right-handed, both have brown eyes - fraternal twins were born. One of them is left-handed with brown eyes, the other is right-handed and has blue eyes. This indicates that the birth of a child with blue eyes indicates a recessive blue eye color. And the fact that they were born left-handed indicates the recessiveness of the better possession of the left hand. You can make the following entries to determine the genotype in parents and children: P AaBb x AaBb, F Abb, aaB, where A is brown eyes, B is right-handed, a is blue eyes, b is left-handed.

Sex Genetics

This type of inheritance is not associated with the X and Y chromosomes, but their combination determines the sex of the organism. The manifestation of dominance and recessiveness of some autosomal genes depends on it. For example, in men, some traits may be dominant, while in women, recessive, and vice versa. When solving problems on the genetics of sex, it must be taken into account that in women in the chromosome set there are 22 pairs of autosomes and two sex chromosomes XX. Men also have 22 pairs of autosomes and sex chromosomes X and Y. The female sex is homogametic, since all 23 pairs of homologous chromosomes will have 22 autosomes and one X chromosome. In men, two types of gametes are formed, namely: 22 + X and 22 + Y, therefore the male sex is heterogametic. This means that the probability of having both girls and boys is the same. Schematically, this can be depicted as follows: P - 22x2 + XX x22x2 + X Y, F - 22x2 + XX 22x2 + X Y.

Depending on the combination of sex chromosomes, during fertilization, the sex of the unborn child is determined. If the egg is fertilized by an X-chromosome sperm, a girl is born, and if it is fertilized by a Y-chromosome, a boy is born.

Tasks on genetics of the exam

When solving the problem on the genetics of the Unified State Examination, one can single out tasks related to issues of dihybrid crossing, inheritance of blood groups and traits, and sex coupling. There are a majority of such problems, but there are also problems of a mixed type, in which it is necessary to consider the inheritance of two pairs of features. Various examples of problems in genetics can be found in the collection, which contains the conditions of the problems and their solution. Here, the general solution of problems is considered, methodological techniques are given that greatly facilitate the solution of such problems.

There are solutions for problems of increased complexity, and at the end of the book are answers for problems that you need to solve on your own. The book also provides an analysis of the typical mistakes made by students. Such a collection can be seen on the Internet, in electronic form. Now you know the general rules of how to solve problems in genetics. This is quite interesting modern science. It will be exciting for both adults and children.

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 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

From seeds obtained by crossing a homozygous plant with yellow seed color with a plant with green seed color, plants were grown, 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

A 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

A 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

A 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)

A 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.

A 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

A task: 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.

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 type, therefore, he received the gene from both his father and mother, therefore his father has a genotype.

	(father)	(mother)

	(was born)

A 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

Determine the number of types of gametes in an organism with a genotype.
Determine the number of types of gametes in an organism with a genotype.
They crossed tall plants with short plants. B - all plants are medium in size. What will be?
They crossed a white rabbit with a black rabbit. All rabbits are black. What will be?
They crossed two rabbits with gray wool. B with black wool, - with gray and white. Determine the genotypes and explain this splitting.
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.
Drosophila 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 trait?
The mother and father have a blood type (both parents are heterozygous). What blood group is possible in children?
The mother has a blood group, the child has a blood group. What blood type is impossible for a father?
The father has the first blood type, the mother has the second. What is the probability of having a child with the first blood type?
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?
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?
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?
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.
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

gamete type.
gamete types.
gamete type.
high, medium and low (incomplete dominance).
black and white.
- black, - white, - grey. 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, - light. Father mother - .

- brown eyes, dark hair
- brown eyes, blonde hair
- blue eyes, dark hair
- blue eyes, blonde hair
- right-handed, - left-handed; Rh positive, Rh negative. Father mother - . Children: (right-handed, Rh positive) and (right-handed, Rh negative).
Father and mother - . In children, a third blood type (probability of birth -) or a first blood group (probability of birth -) is possible.
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.
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, - lefty. Man Woman . Children (healthy boy, right-handed), (healthy girl, carrier, right-handed), (healthy boy, left-handed), (healthy girl, carrier, left-hander).
- 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?
- 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.

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 there is a splitting of signs 3:1 (2784:927 \u003d 3:1), 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 the parental individuals are heterozygous. According to this rule, in this case, the parents must be heterozygous.
Crossing analysis confirms this statement.

First cross scheme

Considering 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. Consequently. 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 crossing. Since when crossing two silkworm breeds that differ from each other in two traits, offspring of the same phenotype were obtained, homozygous individuals were taken when 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 striped 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

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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 (gametes Ab) and aaBB (gametes aB), 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.

Instruction

Certain types of research are used to solve genetic problems. The method of hybridological analysis was developed by G. Mendel. It allows you to identify patterns of inheritance of individual traits during sexual reproduction. The essence of this method is simple: when analyzing certain alternative traits, they can be traced in the offspring. An accurate account of the manifestation of each alternative trait and each individual of the offspring is also carried out.

The basic patterns of inheritance were also developed by Mendel. The scientist deduced three laws. Subsequently, they are so - Mendel's laws. The first is the law of uniformity of hybrids of the first. Take two heterozygous individuals. When crossed, they will give two types of gametes. The offspring of such parents will appear in a ratio of 1:2:1.

Mendel's second law is the law of splitting. at the heart of his assertion that the dominant gene does not always suppress the recessive. In this case, not all individuals among the first generation reproduce the characteristics of their parents - the so-called intermediate nature of inheritance appears. For example, when homozygous plants with red flowers (AA) and white flowers (aa) are crossed, offspring with pink flowers are obtained. Incomplete dominance is quite common. It is also found in some biochemical signs of a person.

The third law and the last one is the law of independent combination of features. For the manifestation of this law, several conditions must be met: there must be no lethal genes, dominance must be complete, the genes must be located on different chromosomes.

The tasks of the genetics of sex 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 time of fertilization. In the sex chromosomes, in addition to the genes that carry information about the sex, there are others that have nothing to do with this. For example, the gene responsible for blood clotting is carried by the female X chromosome. Sex-linked traits are passed from mother to sons and daughters, but from father only to daughters.