Among the tasks in genetics, there are 6 main types found in the exam. 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.

Tasks of the sixth type are tasks of a mixed type. 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.

Below are the theoretical foundations of genetics necessary for successful preparation for task C6, as well as solutions to problems of all types and examples for independent work.

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 F 1

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:

A - yellow color of seeds
a - green color of 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

Seeds obtained by crossing a homozygous yellow seeded plant with a green seeded plant were grown into plants, and F 2 was obtained by self-pollination.

P (F 1)	aa	aa
G	A; a	A; a
F2	AA; Ah; Ah; aa (75% of plants are dominant, 25% are recessive)

The wording of the law: in the offspring obtained from crossing hybrids of the first generation, splitting by phenotype is observed in a ratio of 3: 1, and by genotype - 1: 2: 1.

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.

A - yellow color of seeds, a - green color of seeds,
B - smooth shape, c - wrinkled shape.

Then Mendel grew plants from F 1 seeds and obtained second-generation hybrids by self-pollination.

R	AaVv	AaVv
G	AB, AB, AB, AB	AB, AB, AB, AB
F2	The Punnett grid is used to record and determine genotypes. Gametes AB Av aB av AB AABB AAVv AaBB AaVv Av AAVv aww AaVv aww aB AaBB AaVv aaBB aawww av AaVv aww aawww aavv

In F 2 there was a split into 4 phenotypic classes in a ratio of 9:3:3:1. 9/16 of all seeds had both dominant traits (yellow and smooth), 3/16 - the first dominant and the second recessive (yellow and wrinkled), 3/16 - the first recessive and the second dominant (green and smooth), 1/16 - both recessive trait (green and wrinkled).

When analyzing the inheritance of each pair of traits, the following results are obtained. F 2 contains 12 parts of yellow seeds and 4 parts of green seeds, i.e. 3:1 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 in two or more pairs of alternative traits, genes and their corresponding traits are inherited independently of each other and combined in all possible combinations.

Mendel's third law 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. Both genes appear in F 2, which is possible only if the F 1 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 in a generation, splitting was observed in a ratio of 1: 1, then the original organism contains 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). In the human population, there are three genes (i 0 , I A, I B) encoding erythrocyte antigen proteins that determine people's blood groups. The genotype of each person contains only two genes that determine his blood type: the first group i 0 i 0 ; the second I A i 0 and I A I A; the third I B I B and I B i 0 and the fourth I A I B.

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 - Y and X.

In mammals (including humans), the female sex has a set of sex chromosomes XX, the male sex - XY. 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 the homogametic sex (XX), and females are the heterogametic sex (XY).

The exam includes tasks only for traits linked to the X chromosome. Basically, they relate to two signs of a person: blood clotting (X H - normal; X h - hemophilia), color vision (X D - normal, X d - 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 using hemophilia as an example (a similar picture is observed with color blindness): Х Н Х Н - healthy; X H X h - healthy, but carrier; X h X h - sick. The male sex for these genes is homozygous, tk. The Y-chromosome does not have alleles of these genes: X H Y - healthy; X h Y - 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: 2 n , where n is the number of gene pairs in the heterozygous state. For example, an organism with the AAvvSS genotype does not have genes in the heterozygous state; n \u003d 0, therefore, 2 0 \u003d 1, and it forms one type of gamete (AvC). An organism with the AaBBcc genotype has one pair of genes in the heterozygous state (Aa), i.e. n = 1, therefore 2 1 = 2, and it forms two types of gametes. An organism with the AaBvCs genotype has three pairs of genes in the heterozygous state, i.e. n \u003d 3, therefore, 2 3 \u003d 8, 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 F 1 - 50% white and 50% 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.

For a dihybrid cross

Dominant genes are known

Task: Crossed tomatoes of normal growth with red fruits with dwarf tomatoes with red fruits. In F 1 all plants were of normal growth; 75% - with red fruits and 25% - with yellow. 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 the dominant and recessive genes: A - normal growth, a - dwarfism; B - red fruits, c - yellow fruits.

Let us analyze the inheritance of each trait separately. In F 1, all descendants have normal growth, i.e. splitting on this basis is not observed, so the original forms are homozygous. According to the color of the fruit, a splitting of 3: 1 is observed, therefore, the initial forms are heterozygous.

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 3/8 red saucers, 3/8 red funnels, 1/8 white saucers and 1/8 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 make up 6/8, with white flowers - 2/8, i.e. 3:1 . Therefore, A is red, and white, and the 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 assume that B is saucer-shaped flowers, c is funnel-shaped flowers.

R	AaVv (red flowers, saucer-shaped)	aww (red flowers, funnel-shaped)
G	AB, AB, AB, AB	Av, av
F1	Gametes AB Av aB av Av AAVv AAvv AaVv aww av AaVv aww aawww aavv

3/8 A_B_ - red saucer-shaped flowers,
3/8 A_vv - red funnel-shaped flowers,
1/8 aaBv - white saucer-shaped flowers,
1/8 aavv - 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:

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:

Solving problems of mixed type

Task: A man with brown eyes and 3 blood types married a woman with brown eyes and 1 blood type. They had a blue-eyed child with 1 blood type. Determine the genotypes of all individuals indicated in the problem.

Solution: Brown eyes dominate over blue, so A - brown eyes, a - blue eyes. The child has blue eyes, so his father and mother are heterozygous for this trait. The third blood type may have the genotype I B I B or I B i 0 , the first - only i 0 i 0 . Since the child has the first blood group, therefore, he received the i 0 gene from both his father and mother, therefore his father has the genotype I B i 0.

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 A is right-handed, and left-handed. The male genotype is Aa (because he received the a gene from his left-handed mother), and the female genotype is aa.

A color-blind man has the genotype X d Y, and his wife has X D X D, because her parents were completely healthy.

Tasks for independent solution

1. Determine the number of gamete types in an organism with the AaBBCs genotype.
2. Determine the number of gamete types in an organism with the AaBvX d Y genotype.
3. Determine the number of gamete types in an organism with the genotype aaBBI B i 0 .
4. They crossed tall plants with short plants. In F 1 - all plants are medium in size. What will be F 2 ?
5. They crossed a white rabbit with a black rabbit. In F1, all rabbits are black. What will be F 2 ?
6. They crossed two rabbits with gray wool. In F 1 - 25% with black wool, 50% with gray and 25% with white. Determine the genotypes and explain this splitting.
7. They crossed a black hornless bull with a white horned cow. In F 1, 25% black horns, 25% black horns, 25% white horns and 25% white horns received. Explain this split if black and the absence of horns are dominant traits.
8. 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?
9. A blue-eyed brunette married a brown-eyed blonde. What kind of children can be born if both parents are heterozygous?
10. 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?
11. Mother and father have 3 blood group (both parents are heterozygous). What blood group is possible in children?
12. The mother has 1 blood group, the child has 3 group. What blood type is impossible for a father?
13. 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?
14. A blue-eyed woman with blood group 3 (her parents had a third blood group) married a brown-eyed man with a blood type 2 (his father had blue eyes and a blood type 1). What kind of children can be born?
15. 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?
16. 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. A man with brown eyes and 3 blood types married a woman with brown eyes and 3 blood types. They had a blue-eyed child with 1 blood type. Determine the genotypes of all individuals indicated in the problem.
18. They crossed melons with white oval fruits with plants that had white spherical fruits. The following plants were obtained in the progeny: 3/8 with white oval, 3/8 with white spherical, 1/8 with yellow oval and 1/8 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. 4 types of gametes.
2. 8 types of gametes.
3. 2 types of gametes.
4. 1/4 high, 2/4 medium and 1/4 low (incomplete dominance).
5. 3/4 black and 1/4 white.
6. AA - black, aa - white, Aa - grey. incomplete dominance.
7. Bull: AaBv, cow - aavb. Offspring: AaBv (black hornless), Aavb (black horned), aaBv (white horned), aavb (white hornless).
8. A - red eyes, a - white eyes; B - defective wings, c - normal. Initial forms - AAvv and aaBB, offspring of AaVv.
   Crossing results:
   a) AaBv x AAbv
   • F2
   • Aaaa red eyes, defective wings
   • AABB red eyes, normal wings
   • Aaww red eyes, normal wings

   b) AaBv x aaBV

   • F 2 AaBB red eyes, defective wings
   • Aaaa red eyes, defective wings
   • aaaa white eyes, defective wings
   • aaBB white eyes, defective wings
9. A - brown eyes, a - blue; B - dark hair, c - light. Father is aaBv, mother is Aavb.
   
10. A - right-handed, a - left-handed; B is Rh positive, B is Rh negative. Father is AABB, mother is AABB. Children: 50% AaBv (right-handed, Rh positive) and 50% Aavb (right-handed, Rh negative).
11. Father and mother - I В i 0 . In children, a third blood group is possible (the probability of birth is 75%) or the first blood type (the probability of birth is 25% ).
12. Mother i 0 i 0 , child I В i 0 ; he received the i 0 gene from his mother, and I B from his father. The following blood types are impossible for the father: the second I A I A, the third I B I B, the first i 0 i 0, the fourth I A I B.
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 50%.
14. A - brown eyes, a - blue. Woman aaI B I B, man AaI A i 0 . Children: AaI A I B (brown eyes, fourth group), AaI B i 0 (brown eyes, third group), aaI A I B (blue eyes, fourth group), aaI B i 0 (blue eyes, third group).
15. A is right-handed, a is left-handed. Male AaX h Y, female aaX H X H . Children AaX H Y (healthy boy, right-handed), AaX H X h (healthy girl, carrier, right-handed), aaX H Y (healthy boy, left-handed), aaX H X h (healthy girl, carrier, left-hander).
16. A - red fruits, a - white; B - short-stalked, c - long-stalked.
    Parents: Aavv and aaVv. Offspring: AaBv (red fruits, short-stalked), Aavb (red fruits, long-stalked), aaBv (white fruits, short-stalked), aavb (white fruits, long-stalked).
    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. A - brown eyes, a - blue. Woman AaI B I 0 , man AaI B i 0 . Child: aaI 0 I 0
18. A - white color, a - yellow; B - oval fruits, c - round. Source plants: AaBv and Aavv. Offspring:
    А_Вв - 3/8 with white oval fruits,
    A_vv - 3/8 with white spherical fruits,
    aaBv - 1/8 with yellow oval fruits,
    aavv - 1/8 with yellow spherical fruits.

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

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 in two or more pairs of alternative traits, genes and their corresponding traits are inherited independently of each other and combined in all possible combinations.

Mendel's third law 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 (this is 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 (- norm; - hemophilia), color vision (- norm, - 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 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 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. 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?
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.

Solving problems in genetics using the laws of G. Mendel

One of the tasks of teaching biology is to form students' ideas about the practical significance of biological knowledge as the scientific basis for many modern industries, healthcare, and medicine. Genetics has ample opportunities in the implementation of this task. Important practical tasks of genetics are:

selection of the optimal system of crossing in breeding work and the most effective selection method;

management of the development of hereditary traits;

use of mutagenesis in breeding.

In medicine, the use of genetic knowledge contributes to the development of measures to protect human heredity from the mutagenic effects of environmental factors.

Solving problems in genetics contributes to a better assimilation of the theory. Due to the time limit in the lesson, we consider only solving problems in genetics using the laws of G. Mendel

Lesson objectives:

to get acquainted with the general requirements for the design of the record of the condition of the problem and its solution;

consider different types of problems and examples of their solution;

to consider various ways of solving problems in dihybrid crossing;

Familiarize yourself with the methods of composing various types of tasks.

The main objective of this article is to assist beginner teachers in solving problems and compiling various types of problems using G. Mendel's laws.

STUDY PROCESS

Methods of mastering problem solving techniques

General requirements for the design of records of the conditions of the problem and its solution.

A, V, WITH etc. - genes that determine the manifestation of a dominant trait.
a, b, With etc. - genes that determine the manifestation of a recessive trait.
A– gene for yellow color of pea seeds;
a- the gene for the green color of pea seeds.
The entry is invalid: A- yellow color of pea seeds; a- green color of pea seeds.
Symbol ("mirror of Venus") - used when recording the genotype of the mother (or female);
Symbol ("shield and spear of Mars") - used when recording the genotype of the father (or male).
Crossing is written with an "x".
In crossbreeding schemes, the mother's genotype should be written on the left, the father's genotype on the right.
(For example, in the case of a monohybrid crossing, the entry will look like: AA X aa).
Parents are denoted by the letter R, descendants of the first generation - F1, second - F2 etc.
The letter designations of one or another type of gametes should be written under the designations of the genotypes on the basis of which they are formed.
Place the record of phenotypes under the formulas of the corresponding genotypes.
The numerical ratio of the splitting results should be recorded under the corresponding phenotypes or together with the genotypes.

Consider an example of recording the conditions of the problem and its solution.

Task 1. A blue-eyed youth married a brown-eyed girl whose father had blue eyes. From this marriage, a brown-eyed child was born. What is the child's genotype? (See Table 1 for information on alternative features.)

dominant trait	recessive trait
1. Curly hair (wavy in heterozygotes) 2. Early baldness 3. Not red hair 4. Brown eyes 5. Freckles 6. Dwarfism 7. Polydactyly (extra fingers) 8. Dark hair 9. Rh-positive blood factor 10. Right-handedness	1. Straight hair 2. Norm 3. Red hair 4. Blue or gray eyes 5. No freckles 6. Normal growth 7. Normal number of fingers 8. Blonde hair 9. Rh-negative blood factor 10. Left-handedness
1. Yellow coloration 2. Smooth seed surface 3. Red coloring of the corolla 4. Axillary position of flowers 5. Bloated bean shape 6. Green coloring of beans 7. Tall	1. Green seed color 2. Wrinkled seed surface 3. White corolla 4. Apex position of flowers 5. Flat bean shape 6. Yellow coloring of beans 7. Short stature
1. Round fruit 2. Red color of the fruit 3. Tall	1. Pear-shaped fruits 2. Yellow fruit color 3. Short stature
1. Pea comb 2. Feathered legs	1. Simple comb 2. Unfeathered legs
Cattle
1. Komolost 2. Black wool	1. The presence of horns 2. Red wool
Drosophila
1. Gray body color 2. Normal wings	1. Black coloring of the body 2. Rudimentary wings

Given:

A- brown-eyed gene
a- gene for blue eyes
– aa
– Ah
F1- brown-eyed.

Determine genotype F1

Solution.

Answer: Ah.

The explanation for this task should be as follows.
First, let us briefly write the condition of the problem. According to the table “Alternative traits”, brown eye color is a dominant trait, so the gene that determines this trait is denoted as “ A", and the gene that determines the blue color of the eyes (recessive trait) - as" a».

Given:

A- brown-eyed gene;
a- the gene for blue eyes.

Now let's determine the genotypes of the child's parents. The father is blue-eyed, therefore, in his genotype, both allelic genes that determine eye color are recessive, i.e. his genotype aa.
The child's mother is brown-eyed. The manifestation of this eye color is possible in the following cases.

1. Provided that both allelic genes are dominant.
2. Provided that one of the allelic genes is dominant and the other is recessive. Since the father of the child's mother was blue-eyed, i.e. his genotype aa, then she has one allelic recessive gene. This means that the mother of the child is heterozygous for this trait, her genotype Ah.

In the problem, the phenotype of the child is known - brown-eyed. It is required to know its genotype.

F1- brown-eyed
Genotype F1 – ?

Solution.

Let's write down the genotypes of the parents to the right of the condition of the problem.

R: aa X aa

Knowing the genotypes of the parents, it is possible to determine what types of gametes they form. The mother produces two types of gametes A and a, the father has only one type - a.

R: aa X aa
gametes: A a a

In this marriage, children are possible with two genotypes based on eye color:

aa- brown-eyed and aa- blue-eyed.

The phenotype of the child is known from the condition of the problem: the child is brown-eyed. Therefore, its genotype is Ah.

Answer: brown-eyed child has a genotype Ah.

Note. V F1 another entry is possible:

Skills and abilities necessary for solving problems

I. Before starting to solve problems, students need to firmly master the skills of using alphabetic characters to designate dominant and recessive genes, homo- and heterozygous states of alleles, genotypes of parents and offspring.
For a more solid mastery of these concepts, training exercises can be offered, which are easy to compose using the data in Table. 1–3. And you can use the text of the finished problem, in this case, students are invited to analyze and write down the condition of the problem.

Table 2. Examples of monogenic inheritance of autosomal traits

		Dominant	Recessive
Pumpkin tomatoes Watermelon Onion Silkworm gold fish parrots Mink Person	fruit shape fruit shape fruit coloration Scale coloring Coloring caterpillars The structure of the eyes Plumage coloration Wool coloring Skeleton structure Hearing Vision	Discoid spherical striped Red striped Ordinary Green brown Dwarfism Norm Norm	spherical pear-shaped Smooth yellow Smooth Telescopic blue blue Norm Deafness Blindness

Table 3. Examples of monogenic inheritance of autosomal semi-dominant traits

Exercise 1(according to the table). In cattle, the polled gene (i.e., hornlessness) dominates the horned gene, and the black coat color dominates red, and the genes for both traits are on different chromosomes.

What are the genotypes of cows:

a) black polled;
b) black horned;
c) red horned;
d) red polled?

Given:

A- polled gene;
a- hornedness gene;
V- gene for black coat color;
b- gene for red coat color.

Answer:

a) A _ V _ (i.e. AABB, AaBB, AABb, AaBb)
b) aa V _ (i.e. aaBB, aaBb)
v) aa bb
G) A _ bb(those. AAbb, Aabb)

Exercise 2(from the text of the problem). Plants of red-fruited strawberries, when crossed with each other, always give offspring with red berries, and plants of white-fruited strawberries - with white berries. As a result of crossing both varieties with each other, pink berries are obtained. What offspring arises when hybrid strawberry plants with pink berries are crossed? What offspring will you get if you pollinate a red-fruited strawberry with the pollen of a hybrid strawberry with pink berries?
Write down the condition of the problem and the crossings mentioned in the problem.

Answer:

A+ – gene of redness;
A- white fertility gene;
AA- white-fruited strawberries;
A + A+ - red strawberry;
A + A- strawberries with pink berries.
A + A+ x AA; A + A X A + A;
A + A+ x A + A

II. Another important skill to be mastered is the ability to determine the phenotype by genotype .

Exercise 3 What is the color of pea seeds with the following genotypes: AA, aa, Ah? (See Table 1.)

Answer: yellow; green; yellow.

Exercise 4 What is the shape of the root crop in radish with the following genotypes: AA, Ah, aa? (See Table 3.)

Answer: long; oval; round.

3. Very important learn to write gametes . To calculate the number of different varieties of gametes, the formula 2 n is used, where n is the number of pairs of heterozygous allele states.

For instance:

AA BB CC DD, n = 0; 2 n = 2 0 = 1 (1 kind of gametes) ABCD.

Ah BB CC DD, n = 1; 2 n = 2 1 = 2 (2 kinds of gametes) gametes: ABCD, aBCD.

Ah bb CC DD, n = 2; 2n=4.

Ah bb CC DD, n = 3; 2n=8.

Ah bb CC Dd , n = 4; 2n=16.

For the latter case, consider the notation of gametes. There should be 16 in total.

It is necessary to draw the attention of students to the fact that pairs of genes Ah, bb, CC, Dd are on different chromosomes. During the formation of gametes during meiosis, homologous chromosomes separate, and each germ cell contains a haploid set of chromosomes, that is, each gamete must contain chromosomes with genes A(or a), V (b), WITH (With), D (d). Gametes are not allowed: Ah, bb, CC, Dd or A, a, B, b, C, c, D, d.

Since each pair of traits is inherited independently of the others, then for each pair of alternative traits, genes will be distributed among gametes in the ratio:

Those. in a record of 16 gametes, each gene must be repeated 8 times.

1) Write down the number in order:

4) gene b

1. AB 2. AB 3. AB 4. AB 5. Ab 6. Ab 7. Ab 8. Ab

9. aB 10. aB 11. aB 12. aB 13. ab 14. ab 15. ab 16. ab

5) gene WITH

1. ABC 2. ABC 3. AB 4. AB 5. AbC 6. AbC 7. Ab 8. Ab

9. aBC 10. aBC 11. aB 12. aB 13. abC 14. abC 15. ab 16. ab

6) gene With

1. ABC 2. ABC 3. ABs 4. ABs 4. ABs 5. AbC 6. AbC 7. Abc 8. Abc

9. aBC 10. aBC 11. aBc 12. aBc 13. abC 14. abC 15. abc 16. abc

7) gene D

1. ABCD 2. ABC 3. ABcD 4. ABs 5. AbCD 6. AbC 7. AbcD 8. Abc

9. aBCD 10. aBC 11. aBcD 12. aBc 13. abCD 14. abC 15. abcD 16. abc

8) gene d

1. ABCD 2. ABCd 3. ABcD 4. ABcd 5. AbCD 6. AbCd 7. AbcD 8. Abcd

9. aBCD 10. aBCd 11. aBcD 12. aBcd 13. abCD 14. abCd 15. abcD 16. abcd

This sequence allows you to quickly write down all possible combinations of gene distribution among gametes.

Exercise 5 What types of gametes are formed in plants with genotypes:

1) AABbccDd,
2) AaBbCCD?

Answer:

1) AA bb cc Dd, n = 2; 2n=4
(4 varieties of gametes).

1. ABCD. 2. ABcd. 3. AbcD. 4. Abcd.

2) aa bb CC Dd, n = 3; 2n=8
(8 varieties of gametes).

1. ABCD. 2. ABCd. 3. AbCD.
4. AbCd.5. aBCD. 6. aBCd.
7. abCD. 8. abCd.

We have considered the most complex examples of recording gametes. In the early stages of learning, tasks should be simple. For example, write gametes for genotypes AA, Ah, aa.

To be continued

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) between the internal chromatids of homologous chromosomes, there is an exchange of sections (lucuses). 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.

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.

Related videos

Sources:

• problem solving in biology genetics
• for dihybrid crosses and trait inheritance

All tasks in genetics, as a rule, are reduced to several main types: computational, to determine the genotype and to find out how the trait is inherited. Such tasks may be schematic or illustrated. However, in order to successfully solve any problem, including a genetic one, it is necessary to carefully read its condition. The decision itself is based on the implementation of a number of specific actions.

You will need

• - notebook;
• - a textbook on genetics;
• - pen.

Instruction

First you need to determine the type of proposed task. To do this, it will be necessary to find out how many gene pairs for the development of the proposed traits, which traits are considered. Find out homo- or heterozygous in this case, interbreed with each other, and also whether the inheritance of a particular trait is associated with the sex chromosomes.

Find out which of the features proposed for study is (weak) and which is dominant (strong). At the same time, when solving a genetic problem, it is necessary to start from the premise that the dominant trait in the offspring will always manifest itself phenotypically.

Determine the number and type of gametes (sex). It should be borne in mind that gametes can only be haploid. Accordingly, the distribution of chromosomes during their division occurs evenly: each of the gametes will contain only one chromosome taken from a homologous pair. As a result, the offspring receives a "half" set of chromosomes from each of their own.

Make a schematic record of the conditions of the genetic problem in a notebook. In this case, the dominant traits for a homozygous organism under study are designated as a combination of AA, for a heterozygous - Aa. An indeterminate genotype is designated A_. A recessive trait is written as a combination of aa.

Write down the phenotypes and genotypes of individuals crossed according to the condition of the problem. Then, focusing on point 3 (definition of gamete types), write down the phenotypes and genotypes of the offspring obtained as a result of crossing.

Analyze the results and write down this numerical ratio. This will be the answer to the genetic problem.

Related videos

Useful advice

In many similar tasks, the genotype of individuals proposed for crossing is not specified. That is why it is so important to be able to independently determine the genotype of parents by the phenotype or genotype of their offspring.

In the study of genetics, much attention is paid to problems whose solution must be found using the laws of gene inheritance. To most students of the natural sciences, solving problems in genetics seems to be one of the most difficult things in biology. However, it is found by a simple algorithm.

You will need

• - textbook.

Instruction

To begin with, carefully read the problem and write down a schematic condition using special characters. Indicate what genotypes the parents have and what phenotype corresponds to them. Write down what kind of children came out in the first and second generations.

Note which gene is dominant and which is recessive, if it is in the condition. If splitting is given in the problem, also indicate it in the schematic notation. For simple problems, sometimes it is enough to write down the condition in order to understand the solution to the problem.

To successfully solve the problem, you need to understand which section it belongs to: monohybrid, dihybrid or polyhybrid crossing, sex-linked inheritance, or the trait is inherited during the interaction of genes. To do this, calculate what splitting of the genotype or phenotype is observed in the offspring in the first generation. The condition may indicate the exact number of individuals with each genotype or phenotype, or the percentage of each genotype (phenotype) of the total. These data must be reduced to prime numbers.

Pay attention to whether the offspring have signs depending on the sex.

Each type of crossing is characterized by its own special splitting according to the genotype and phenotype. All this data is contained in the textbook, and it will be convenient for you to write these formulas on a separate sheet and use them when solving problems.

Now that you have discovered the splitting that transmits hereditary traits in your problem, you can find out the genotypes and phenotypes of all individuals in the offspring, as well as the genotypes and phenotypes of the parents involved in the crossing.

Write down the received data in the answer.

All tasks in biology are divided into tasks in molecular biology and tasks in genetics. In molecular biology, there are several topics that have problems: proteins, nucleic acids, DNA code, and energy metabolism.

Instruction

Solve problems on the topic "Proteins" using the following formula: m(min) \u003d a / b * 100%, where m (min) is the molecular weight, a is the atomic or molecular weight of the component, b is the percentage of the component. The average molecular weight of one acid residue is 120.

Calculate the required values ​​​​on the topic “Nucleic acids”, adhering to Chargaff: 1. The amount of adenine is equal to the amount of thymine, and guanine is equal to cytosine;
2. The number of purine bases is equal to the number of pyrimidine bases, i.e. A + G \u003d T + C. In the chain of a DNA molecule, the distance between nucleotides is 0.34 nm. The relative molecular weight of one nucleotide is 345.

Solve problems on the topic “DNA Code” using a special table of genetic codes. Thanks to her, you will find out which acid encodes a particular genetic code.

Calculate the answer you need for tasks on the topic "Energy Exchange" using the reaction equation. One of the most common is: С6Н12О6 + 6О2 → 6СО2 + 6Н2О.

Search for genetics using a special algorithm. First, determine which genes are dominant (A, B) and which are recessive (a, b). A gene is called dominant, the trait of which manifests itself both in the homozygous (AA, aa) and in the heterozygous state (Aa, Bb). A gene is called recessive, the sign of which is manifested only when the same genes meet, i.e. in a homozygous state. For example, yellow seeded pea plants are crossed with green seeded pea plants. The resulting pea plants all had yellow seeds. Obviously, the yellow color of the seeds is the dominant trait. Write the solution to this problem as follows: A - the gene responsible for the yellow color of the seeds, and - the gene responsible for the green color of the seeds. P: AA x aa
G: A
F1: AaThere are tasks of this type with several features, then one feature is designated A or a, and the second B or b.

The study of genetics is accompanied by problem solving. They clearly show the operation of the law of inheritance of genes. Most students find these problems incredibly difficult to solve. But, knowing the solution algorithm, you can easily cope with them.

Instruction

Two main types can be distinguished. In the first type of tasks, the genotypes of the parents are known. It is necessary to determine the genotypes of the offspring. First determine which allele is dominant. Find the allele. Write down the genotypes of the parents. Write down all possible types of gametes. Connect . Define split.

In tasks of the second type, the opposite is true. Here, splitting in the offspring is known. It is required to determine the genotypes of the parents. Find, just as in tasks of the first type, which of the alleles is dominant, which is recessive. Determine the possible types of gametes. Based on them, determine the genotypes of the parents.

To solve the problem correctly, read it carefully and analyze the condition. To determine the type of problem, find out how many feature pairs are considered in the problem. Notice also how many pairs of genes control the development of traits. It is important to find out if they are homozygous or crossbreeding, what type of crossbreeding. Determine whether genes are inherited independently or linked, how many genotypes are formed in the offspring, and whether inheritance is sex-linked.