Dihybrid crossing. Examples of solving typical tasks

Objective 1. In humans, complex forms of myopia dominate normal vision, brown eye color - over blue. A brown-eyed nearsighted man whose mother had blue eyes and normal vision married a blue-eyed woman with normal vision. What is the% probability of having a baby with signs of a mother?

Decision

Gene Sign

A development of myopia

a normal vision

B Brown eyes

b blue eyes

P ♀ aabb x ♂ AaBb

G ab, AB, Ab aB, ab

F 1 AaBb; Aabb; aaBb; aabb

Answer: a child with the aabb genotype has blue eyes and normal vision. The probability of having a child with these signs is 25%.

Problem 2... In humans, red hair color dominates over light brown, and freckles over their absence. A heterozygous red-haired man without freckles married a fair-haired woman with freckles. Determine in% the probability of having a red-haired child with freckles.

Decision

Gene Sign

A red hair

a blonde hair

Freckles

b lack of freckles

P ♀ Aabb x ♂ aaBB

F 1 AaBb; aaBb

A red-haired child with freckles has the AaBb genotype. The probability of having such a child is 50%.

Answer: The odds of having a redhead with freckles are 50%.

Problem 3... A heterozygous woman with a normal hand and freckles married a six-fingered heterozygous man who does not have freckles. What is the probability of having a baby with a normal hand and without freckles?

Decision

Gene Sign

A six-fingered (polydactyly),

anormal brush

Bthe presence of freckles

bno freckles

P ♀ aaBb x ♂ Aаbb

G aB, ab, Ab, ab

F 1 AaBb; Aabb; aaBb; aabb

Answer: The probability of having a baby with the aabb genotype (with a normal hand, no freckles) is 25%.

Task 4... The genes for cataract predisposition and red hair are on different pairs of chromosomes. A red-haired woman with normal vision married a fair-haired man with cataracts. What phenotypes can they have children with if the man's mother has the same phenotype as his wife?

Decision

Gene Sign

Ablonde hair,

ared hair

Bdevelopment of cataracts

bnormal vision

P ♀ aabb x ♂ AaBb

G ab, AB, Ab, aB, ab

F 1 AaBb; Aabb; aaBb; aabb

Answer: phenotypes of children - fair-haired with cataracts (AaBb); blonde without cataracts (Aabb); redhead with cataracts (aaBb); redhead without cataracts (aabb).

Objective 5. What is the percentage probability of having a child with diabetes mellitus if both parents are carriers of the recessive gene for diabetes mellitus? At the same time, the mother's blood Rh factor is positive, and the father's is negative. Both parents are homozygous for the gene that determines the development of the Rh factor. Blood, with what Rh factor will the children of this married couple?

Decision

Gene Sign

Anormal carbohydrate metabolism

a development of diabetes mellitus

Rh + Rh positive blood

rh -rh negative blood.

P♀ AaRh + Rh + x ♂ Aarh - rh -

G ARh +, aRh +, Arh - , arh -

F 1 AARh + rh - ; AaRh + rh - ; AaRh + rh - ; aaRh + rh -

Answer: the probability of having a child with diabetes mellitus is 25%, all children in this family will have a positive Rh factor.

Problem 6... Normal growth in oats dominates over gigantism, early maturity over late maturity. The genes for both traits are on different pairs of chromosomes. What percentage of late-maturing plants of normal growth can be expected from crossing plants heterozygous for both traits?

Decision

P ♀ AaBb x ♂ AaBb

G AB, Ab, AB, Ab,

A plant with red fruits forms gametes carrying dominant alleles AB, and a plant with yellow fruits forms gametes carrying recessive alleles aw... The combination of these gametes leads to the formation of a diheterozygote AaBvsince genes AND and IN dominant, then all first generation hybrids will have red and smooth fruits.

We cross plants with red and smooth fruits from generation F 1with a plant with yellow and pubescent fruits (Fig. 2). Let's define the genotype and phenotype of the offspring.

Figure: 2. Crossing scheme ()

One of the parents is a diheterozygote, his genotype AaBv, the second parent is homozygous for recessive alleles, its genotype is aavv... A diheterozygous organism produces the following types of gametes: AB, Av, aB, aw; homozygous organism - gametes of the same type: aw... The result is four genotypic classes: AaBv, Aavb, aaBv, aavv and four phenotypic classes: red smooth, red pubescent, yellow smooth, yellow pubescent.

Splitting for each of the signs: 1: 1 for fruit color, 1: 1 for fruit skin.

This is a typical analytical crossing that allows you to determine the genotype of an individual with a dominant phenotype. A dihybrid cross is two independently running monohybrid crosses, the results of which overlap. The described mechanism of inheritance in a dihybrid crossing refers to traits whose genes are located in different pairs of non-homologous chromosomes, that is, genes responsible for the color of tomato fruits are located in one pair of chromosomes, and genes responsible for the smoothness or pubescence of the fruit skin are located in the other pair of chromosomes.

The crossing of two pea plants grown from yellow and smooth seeds yielded 264 yellow smooth, 61 yellow wrinkled, 78 green smooth, 29 green wrinkled seeds. Determine to which cross the observed ratio of phenotypic classes belongs.

In the condition, splitting from crossing is given, four phenotypic classes were obtained with the following splitting 9: 3: 3: 1, and this indicates that two diheterozygous plants with the following genotype were crossed: AaBv and AaBv (Fig. 3).

Figure: 3 Crossing scheme for problem 2 ()

If we construct a Punnett lattice, in which we write down gametes horizontally and vertically, and zygotes obtained by merging gametes in squares, we get four phenotypic classes with the splitting indicated in the problem (Fig. 4).

Figure: 4. Punnet lattice to problem 2 ()

Incomplete dominance in one of the characteristics. In the snapdragon plant, the red color of the flowers does not completely suppress the white color, the combination of the dominant and recessive alleles causes the pink color of the flowers. The normal flower shape dominates the elongated and pyloric flower shape (Fig. 5).

Figure: 5. Crossing the snapdragon ()

Homozygous plants with normal white flowers and a homozygous plant with elongated red flowers were crossed. It is necessary to determine the genotype and phenotype of the offspring.

The task:

AND - red color is a dominant feature

and - white color is a recessive sign

IN - normal form - dominant feature

in- pyloric form - a recessive sign

aaBB- genotype of white color and normal flower shape

AAvv - genotype of red pyloric flowers

They produce gametes of the same type, in the first case, gametes carrying alleles aB, in the second case - Av... The combination of these gametes leads to the emergence of a diheterozygote, which has the genotype AaBv - all hybrids of the first generation will have a pink color and normal flower shape (Fig. 6).

Figure: 6. Crossing scheme for problem 3 ()

We cross the first generation hybrids to determine the color and shape of the flower in the generation F 2 with incomplete dominance in color.

Genotypes of parental organisms - AaBv and AvVv,

hybrids form gametes of four types: AB, Av, aB, aw (Fig. 7).

Figure: 7. Crossing scheme for hybrids of the first generation, task 3 ()

When analyzing the obtained offspring, we can say that we did not succeed in the traditional splitting according to the phenotype of 9: 3 and 3: 1, since the plants have incomplete dominance in the color of flowers (Fig. 8).

Figure: 8. Punnet table for problem 3 ()

Of the 16 plants: three red normal, six pink normal, one red pyloric, two pink pyloric, three white normal and one white pyloric.

We have considered examples of solving problems for dihybrid crossing.

In humans, brown eye color dominates over blue, and the ability to better control the right hand dominates over left-handedness.

Task 4

The brown-eyed right-hander married a blue-eyed left-hander, they had two children - a blue-eyed right-hander and a blue-eyed left-hander. Determine the genotype of the mother.

Let's write down the condition of the problem:

AND - Brown eyes

and- blue eyes

IN - right-handedness

in - left-handedness

aavv - father's genotype, he is homozygous for recessive alleles of two genes

AND - ? IN -? - the mother's genotype has two dominant genes and can theoretically have

genotypes: AABB, AaBB, AABv, AaBv.

F 1 - aavv, aaB - ?

If there is a genotype AABB the mother would not show any splitting in the offspring: all children would be brown-eyed right-handed and would have the genotype AaBv, since the father forms gametes of the same type aw(Fig. 9).

Figure: 9. Crossing scheme for problem 4 ()

Two children have blue eyes, which means the mother is heterozygous for eye color Aa, besides this, one of the children is left-handed - this suggests that the mother has a recessive gene inresponsible for left-handedness, that is, the mother is a typical diheterozygote. The crossing scheme and possible children from this marriage are shown in Fig. ten.

Figure: 10. Crossing scheme and possible children from marriage ()

A trihybrid crossing is a crossing in which the parental organisms differ from each other in three pairs of alternative characters.

Example: Crossing peas with smooth yellow seeds and purple flowers with green wrinkled seeds and white flowers.

In trihybrid plants dominant traits will appear: yellow color and smooth seed shape with a purple flower color (Fig. 11).

Figure: 11. Scheme of trihybrid crossing ()

Trihybrid plants, as a result of independent gene cleavage, produce

eight types of gametes - female and male, combining, they will give in F 264 combinations, 27 genotypes and 8 phenotypes.

List of references

1. Mamontov S.G., Zakharov V.B., Agafonova I.B., Sonin N.I. Biology grade 11. General biology. Profile level. - 5th edition, stereotyped. - Bustard, 2010.
2. Belyaev D.K. General biology. A basic level of. - 11th edition, stereotyped. - M .: Education, 2012.
3. Pasechnik V.V., Kamensky A.A., Kriksunov E.A. General biology, grades 10-11. - M .: Bustard, 2005.
4. Agafonova I.B., Zakharova E.T., Sivoglazov V.I. Biology 10-11 grade. General biology. A basic level of. - 6th ed., Add. - Bustard, 2010.

1. Biorepet-ufa.ru ().
2. Kakprosto.ru ().
3. Genetika.aiq.ru ().

Homework

1. Define dihybrid crossing.
2. Write the possible types of gametes produced by organisms with the following genotypes: AABB, CcDD.
3. Give a definition of trihybrid crossing.

Genetics, its tasks. Heredity and variability are the properties of organisms. Genetic methods. Basic genetic concepts and symbols. Chromosomal theory of heredity. Modern understanding of the gene and genome

Genetics, its tasks

The successes of natural science and cell biology in the 18th-19th centuries allowed a number of scientists to make assumptions about the existence of certain hereditary factors that determine, for example, the development of hereditary diseases, but these assumptions were not supported by appropriate evidence. Even the theory of intracellular pangenesis, formulated by H. de Vries in 1889, which assumed the existence in the cell nucleus of certain "pangens" that determine the hereditary inclinations of the organism, and the exit into the protoplasm of only those of them that determine the type of cell, could not change the situation, as well as theory of "germplasm" A. Weismann, according to which acquired in the process of ontogeny, traits are not inherited.

Only the works of the Czech researcher G. Mendel (1822-1884) became the founding stone of modern genetics. However, despite the fact that his works were cited in scientific publications, contemporaries did not pay attention to them. And only the rediscovery of the patterns of independent inheritance by three scientists at once - E. Cermak, K. Correns and H. de Vries - forced the scientific community to turn to the origins of genetics.

Genetics Is a science that studies the laws of heredity and variability and methods of managing them.

The tasks of genetics at the present stage are the study of the qualitative and quantitative characteristics of the hereditary material, the analysis of the structure and functioning of the genotype, the deciphering of the fine structure of the gene and methods of regulation of gene activity, the search for genes that cause the development of hereditary human diseases and methods of their "correction", the creation of a new generation of drugs by type DNA vaccines, the construction of organisms with new properties using the means of genetic and cellular engineering that could produce drugs and food products necessary for a person, as well as a complete decoding of the human genome.

Heredity and variability - properties of organisms

Heredity - This is the ability of organisms to transmit their traits and properties in a series of generations.

Variability - the property of organisms to acquire new characteristics during life.

Signs - any morphological, physiological, biochemical and other features of organisms, by which some of them differ from others, for example, eye color. Properties any functional features of organisms are called, which are based on a certain structural feature or a group of elementary features.

The traits of organisms can be divided into quality and quantitative... Qualitative signs have two or three contrasting manifestations, which are called alternative signs, for example, blue and brown eye color, while quantitative (milk yield of cows, yield of wheat) do not have clear differences.

The material carrier of heredity is DNA. In eukaryotes, two types of inheritance are distinguished: genotypic and cytoplasmic... Carriers of genotypic heredity are localized in the nucleus and further we will talk about it, and the carriers of cytoplasmic heredity are circular DNA molecules located in mitochondria and plastids. Cytoplasmic inheritance is transmitted mainly with the egg, therefore it is also called maternal.

A small number of genes are localized in the mitochondria of human cells, however, their change can have a significant impact on the development of the body, for example, lead to the development of blindness or a gradual decrease in mobility. Plastids play an equally important role in plant life. So, in some parts of the leaf, chlorophyll-free cells may be present, which leads, on the one hand, to a decrease in the productivity of the plant, and on the other, such variegated organisms are valued in decorative gardening. Such specimens are reproduced mainly asexually, since during sexual reproduction, ordinary green plants are more often obtained.

Genetic methods

1. The hybridological method, or the method of crossing, consists in the selection of parental individuals and analysis of the offspring. At the same time, the genotype of an organism is judged by the phenotypic manifestations of genes in the offspring obtained with a certain crossing pattern. This is the oldest informative method of genetics, which was most fully used for the first time by G. Mendel in combination with the statistical method. This method is not applicable in human genetics for ethical reasons.

2. The cytogenetic method is based on the study of the karyotype: the number, shape and size of the body's chromosomes. The study of these features makes it possible to identify various developmental pathologies.

3. The biochemical method allows you to determine the content of various substances in the body, especially their excess or deficiency, as well as the activity of a number of enzymes.

4. Molecular genetic methods are aimed at identifying variations in the structure and deciphering the primary nucleotide sequence of the studied DNA regions. They make it possible to identify genes of hereditary diseases even in embryos, establish paternity, etc.

5. Population-statistical method allows you to determine the genetic composition of the population, the frequency of certain genes and genotypes, genetic load, as well as outline the prospects for the development of the population.

6. The method of hybridization of somatic cells in culture makes it possible to determine the localization of certain genes in chromosomes during the fusion of cells of different organisms, for example, a mouse and a hamster, a mouse and a human, etc.

Basic genetic concepts and symbols

Gene - This is a section of a DNA molecule, or chromosome, which carries information about a certain trait or property of an organism.

Some genes can influence the manifestation of several traits at once. This phenomenon is called pleiotropy... For example, a gene that causes the development of a hereditary disease of arachnodactyly (spider fingers) also causes a curvature of the lens, pathology of many internal organs.

Each gene occupies a strictly defined place in the chromosome - locus... Since in the somatic cells of most eukaryotic organisms, the chromosomes are paired (homologous), then in each of the paired chromosomes there is one copy of the gene responsible for a certain trait. Such genes are called allelic.

Allelic genes most often exist in two variants - dominant and recessive. Dominant is called an allele that manifests itself regardless of which gene is on the other chromosome, and suppresses the development of a trait encoded by a recessive gene. Dominant alleles are usually indicated by uppercase letters of the Latin alphabet (A, B, C, etc.), and recessive ones by lowercase (a, b, c, etc.). Recessive alleles can only appear if they occupy loci on both paired chromosomes.

An organism that has the same alleles in both homologous chromosomes is called homozygous for a given gene, or homozygote (AA, aa, AABB, aabb, etc.), and an organism in which there are different variants of the gene in both homologous chromosomes - dominant and recessive - is called heterozygous for a given gene, or heterozygote (Aa, AaBb, etc.).

A number of genes can have three or more structural variants, for example, blood groups according to the AB0 system are encoded by three alleles - I A, I B, i. This phenomenon is called multiple allelism. However, even in this case, each chromosome from a pair carries only one allele, that is, all three variants of a gene in one organism cannot be represented.

Genome - a set of genes characteristic of a haploid set of chromosomes.

Genotype - a set of genes characteristic of a diploid set of chromosomes.

Phenotype - a set of signs and properties of an organism, which is the result of the interaction of the genotype and the environment.

Since organisms differ in many traits, it is possible to establish the patterns of their inheritance only by analyzing two or more traits in the offspring. Crossbreeding, in which inheritance is considered and an accurate quantitative accounting of offspring is carried out for one pair of alternative characters, is called monohybridm, in two pairs - dihybrid, for most of the signs - polyhybrid.

By the phenotype of an individual, it is far from always possible to establish its genotype, since both an organism homozygous for the dominant gene (AA) and a heterozygous one (Aa) will have a dominant allele in the phenotype. Therefore, to check the genotype of an organism with cross fertilization, they use analyzing cross - crossing, in which an organism with a dominant trait is crossed with a homozygous for a recessive gene. In this case, an organism homozygous for the dominant gene will not split in the offspring, while in the offspring of heterozygous individuals there is an equal number of individuals with dominant and recessive traits.

To record crossing schemes, the following conventions are most often used:

P (from lat. parent - parents) - parental organisms;

$ ♀ $ (alchemical sign of Venus - a mirror with a handle) - a mother;

$ ♂ $ (alchemical sign of Mars - shield and spear) - father;

$ × $ - crossing sign;

F 1, F 2, F 3, etc. - hybrids of the first, second, third and subsequent generations;

F a - offspring from the analyzing cross.

Chromosomal theory of heredity

The founder of genetics G. Mendel, as well as his closest followers, did not have the slightest idea about the material basis of hereditary inclinations, or genes. However, already in 1902-1903, the German biologist T. Boveri and the American student W. Setton independently suggested that the behavior of chromosomes during cell maturation and fertilization makes it possible to explain the splitting of hereditary factors according to Mendel, i.e., in their opinion, genes must be located on chromosomes. These assumptions became the cornerstone of the chromosomal theory of heredity.

In 1906, English geneticists W. Batson and R. Pennett discovered a violation of Mendelian cleavage when crossing sweet peas, and their compatriot L. Doncaster discovered sex-linked inheritance in experiments with the goose moth butterfly. The results of these experiments clearly contradicted Mendelian ones, but considering that by that time it was already known that the number of known characters for experimental objects was much higher than the number of chromosomes, and this suggested that each chromosome carries more than one gene, and the genes of one chromosomes are inherited together.

In 1910, T. Morgan's group began experiments on a new experimental facility - the fruit fly Drosophila. The results of these experiments made it possible by the middle of the 20s of the XX century to formulate the main provisions of the chromosomal theory of heredity, to determine the order of arrangement of genes in chromosomes and the distance between them, that is, to draw up the first maps of chromosomes.

The main provisions of the chromosomal theory of heredity:

1. Genes are located on chromosomes. Genes of one chromosome are inherited jointly, or linked, and are called clutch group... The number of linkage groups is numerically equal to the haploid set of chromosomes.
2. Each gene occupies a strictly defined place in the chromosome - a locus.
3. Genes in chromosomes are arranged linearly.
4. Disruption of gene linkage occurs only as a result of crossing over.
5. The distance between genes on a chromosome is proportional to the percentage of crossing over between them.
6. Independent inheritance is characteristic only for genes of non-homologous chromosomes.

Modern understanding of the gene and genome

In the early 40s of the twentieth century, J. Beadle and E. Tatum, analyzing the results of genetic studies carried out on the neurospore fungus, came to the conclusion that each gene controls the synthesis of an enzyme, and formulated the principle "one gene - one enzyme" ...

However, already in 1961 F. Jacob, J. L. Monod and A. Lvov managed to decipher the structure of the E. coli gene and study the regulation of its activity. For this discovery, he was awarded the Nobel Prize in Physiology or Medicine in 1965.

In the process of research, in addition to structural genes that control the development of certain traits, they were able to identify regulatory ones, the main function of which is the manifestation of traits encoded by other genes.

The structure of the prokaryotic gene. The structural gene of prokaryotes has a complex structure, since it includes regulatory regions and coding sequences. Regulatory sites include a promoter, operator, and terminator. Promoter The name of the region of the gene to which the RNA polymerase enzyme is attached, which ensures the synthesis of mRNA during the transcription process. FROM operatorlocated between the promoter and the structural sequence can bind repressor protein, which does not allow RNA polymerase to start reading hereditary information from the coding sequence, and only its removal allows transcription to start. The structure of the repressor is usually encoded in a regulatory gene located in another region of the chromosome. The reading of information ends at a section of the gene called terminator.

Coding sequence a structural gene contains information about the amino acid sequence in the corresponding protein. The coding sequence in prokaryotes is called cistron, and the set of coding and regulatory regions of the prokaryotic gene is operon... In general, prokaryotes, which include E. coli, have a relatively small number of genes located on a single ring chromosome.

The cytoplasm of prokaryotes can also contain additional small circular or unclosed DNA molecules called plasmids. Plasmids are able to integrate into chromosomes and be transmitted from one cell to another. They can carry information about sex characteristics, pathogenicity, and antibiotic resistance.

The structure of the eukaryotic gene. Unlike prokaryotes, eukaryotic genes do not have an operon structure, since they do not contain an operator, and each structural gene is accompanied only by a promoter and terminator. In addition, in the genes of eukaryotes, significant regions ( exons) alternate with insignificant ( introns), which are completely rewritten to mRNA and then cut out during their maturation. The biological role of introns is to reduce the likelihood of mutations in significant sites. The regulation of genes in eukaryotes is much more complex than that described for prokaryotes.

The human genome. Each human cell contains about 2 m of DNA in 46 chromosomes, tightly packed in a double helix, which consists of approximately $ 3.2 × $ 10 9 nucleotide pairs, which provides about 10.19 billion possible unique combinations. By the end of the 80s of the twentieth century, the location of about 1500 human genes was known, but their total number was estimated at about 100 thousand, since only hereditary diseases in humans have about 10 thousand, not to mention the number of various proteins contained in cells ...

In 1988, the international project "Human Genome" was launched, which ended by the beginning of the XXI century complete transcript nucleotide sequences. He made it possible to understand that two different people have 99.9% similar nucleotide sequences, and only the remaining 0.1% determine our individuality. In total, about 30-40 thousand structural genes were discovered, but then their number was reduced to 25-30 thousand. Among these genes there are not only unique, but also repeated hundreds and thousands of times. Nevertheless, these genes encode a much larger number of proteins, for example, tens of thousands of protective proteins - immunoglobulins.

97% of our genome is genetic "garbage" that exists only because it is able to reproduce well (RNAs that are transcribed in these regions never leave the nucleus). For example, among our genes there are not only "human" genes, but also 60% of genes similar to genes of the Drosophila fly, and up to 99% of genes have in common with chimpanzees.

In parallel with the decoding of the genome, mapping of chromosomes took place, as a result of which it was possible not only to discover, but also to determine the location of some genes responsible for the development of hereditary diseases, as well as target genes of drugs.

Deciphering the human genome has not yet yielded a direct effect, since we received a kind of instruction for assembling such a complex organism as a human, but did not learn how to make it or even correct errors in it. Nevertheless, the era of molecular medicine is already on the verge, the development of so-called gene drugs is underway all over the world, which can block, remove or even replace pathological genes in living people, and not just in a fertilized egg.

It should not be forgotten that in eukaryotic cells, DNA is contained not only in the nucleus, but also in mitochondria and plastids. Unlike the nuclear genome, the organization of mitochondrial and plastid genes has much in common with the organization of the prokaryotic genome. Despite the fact that these organelles carry less than 1% of the hereditary information of the cell and do not even encode a complete set of proteins necessary for their own functioning, they can significantly affect some of the characteristics of the organism. So, variegation in plants of chlorophytum, ivy and others is inherited by a small number of descendants, even when two variegated plants are crossed. This is due to the fact that plastids and mitochondria are transmitted for the most part with the cytoplasm of the egg, therefore this inheritance is called maternal, or cytoplasmic, in contrast to genotypic, which is localized in the nucleus.

Municipal state educational institution Lyceum No. 4

rossosh of the Rossoshansk municipal district of the Voronezh region.

Methodical development in biology to help students taking the exam.

"Problems in genetics" for grade 11

biology teacher of the highest qualification category

2016

I. Monohybrid crossing problems (complete and incomplete dominance)

1. In rabbits, the gray color of the coat dominates over black. A homozygous gray rabbit was crossed with a black rabbit. Identify the phenotypes and genotypes of the rabbits?
2. In guinea pigs, the black color of the coat dominates over the white. Two heterozygous male and female were crossed. What will the first generation hybrids be like? What law is manifested in this inheritance?
3. When two white pumpkins were crossed in the first generation, ¾ of the plants were white and ¼ were yellow. What are the genotypes of the parents if white dominates over yellow? 4. Black Cow Night brought a red calf. The red cow Zorka has a black calf. These cows are from the same herd, in which there is one bull. What are the genotypes of all animals? Consider different options. ( The black gene is dominant.)
5. How many dwarf pea plants can be expected when sowing 1200 seeds obtained by self-pollination of tall heterozygous pea plants? (Seed germination is 80%).

6. The fruit of a watermelon can be green or striped. All watermelons obtained by crossing plants with green and striped fruits had only a green fruit rind. What color of watermelon fruit can be in F 2?

7.In snapdragons, plants with wide leaves, when crossed, always produce offspring with the same leaves, and when a narrow-leaved plant is crossed with a broad-leaved plant, plants with leaves of intermediate width appear. What will be the offspring from crossing two individuals with leaves of intermediate width.

II

1. In humans, glaucoma is inherited as an autosomal recessive trait (a), and Marfan's syndrome, accompanied by an abnormality in the development of connective tissue, is inherited as an autosomal dominant trait (B). Genes are found in different pairs of autosomes. One of the spouses suffers from glaucoma and did not have ancestors with Marfan's syndrome, and the second is diheterozygous for these characteristics. Determine the genotypes of the parents, possible genotypes and phenotypes of children, the probability of birth healthy child... Make a scheme for solving the problem. What is the law of heredity manifested in this case?

2. Low-growing (dwarf) tomato plants with ribbed fruits and plants of normal height with smooth fruits were crossed. In the offspring, two phenotypic groups of plants were obtained: undersized with smooth fruits and normal height with smooth fruits. When crossing undersized tomato plants with ribbed fruits with plants having a normal stem height and ribbed fruits, all the offspring had a normal stem height and ribbed fruits. Make crossings schemes. Determine the genotypes of the parents and offspring of tomato plants in two crosses. What is the law of heredity manifested in this case?

3. When crossing horned red cows with hornless black bulls, calves of two phenotypic groups were born: horned black and hornless black. Upon further crossing of the same horned red cows with other hornless black bulls, the offspring included hornless red and hornless black bulls. Write schemes for solving the problem. Determine the genotypes of the parents and offspring in two crosses. What is the law of heredity manifested in this case?

4. Fruits of tomatoes can be red and yellow, bare and pubescent. Genes for yellow coloration and pubescence are known to be recessive. Of the tomatoes harvested on the collective farm, there were 36 tons of red pubescent and 12 tons of red pubescent. How many (approximately) in a collective farm harvest can there be yellow fluffy tomatoes if the source material was heterozygous?

5. When crossing a variegated crested (B) hen with the same rooster, eight chickens were obtained: four variegated crested chickens, two white (a) crested and two black crested chickens. Make a scheme for solving the problem. Determine the genotypes of parents, offspring, explain the character of inheritance of traits and the appearance of individuals with a variegated color. What laws of heredity are manifested in this case?

6. In mice, black coat color dominates over brown (a). The long tail is determined by the dominant gene (B) and develops only in the homozygous state; the short tail develops in heterozygotes. Recessive genes that determine the length of the tail in a homozygous state cause the death of embryos. The genes of the two traits are not linked.When mating a female mouse with black hair and a short tail and a male with brown hair and a short tail, the following offspring were obtained: black mice with a long tail, black with a short tail, brown mice with a long tail and brown mice with a short tail. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, the ratio of phenotypes and genotypes of the offspring, the probability of embryo death. What is the law of heredity manifested in this case? Justify the answer.
7. We crossed a plant of whiskered white-fruited strawberries with beanless red-fruited plants (B). All hybrids turned out to be whiskered pink-fruited. When analyzing the crossing of F 1 hybrids, phenotypic cleavage occurred in the offspring. Make a scheme for solving the problem. Determine the genotypes of the parents, first generation hybrids, as well as the genotypes and phenotypes of the offspring in the analyzing crossing (F 2). Determine the nature of inheritance of the trait of color of the fetus. What laws of heredity are manifested in these cases?
8. When crossing phlox plants with white flowers and a funnel-shaped corolla with a plant with cream flowers and flat corollas, 78 offspring were obtained, of which 38 form white flowers with flat corollas, and 40 - cream flowers with flat corollas. When phloxes with white flowers and funnel-shaped corollas were crossed with a plant with cream flowers and flat corollas, phloxes of two phenotypic groups were obtained: white with funnel-shaped corollas and white with flat corollas. Make a diagram of the two crosses. Determine the genotypes of the parents and offspring in two crosses. What is the law of heredity manifested in this case?

9. There are two types of hereditary blindness, each of which is determined by recessive gene alleys (a or b). Both alleles are on different pairs of homologous chromosomes. What is the probability of the birth of a blind grandchild in a family in which maternal and paternal grandmothers are dihomozygous and suffer different kinds blindness, and both grandfathers can see well (do not have recessive genes). Make a scheme for solving the problem. Identify the genotypes and phenotypes of grandparents, their children and possible grandchildren.

III... Tasks for the inheritance of blood groups of the AB0 system

The boy has group I, his sister has group IV. What are the genotypes of the parents?

The father has blood type IV, the mother has I. Can a child inherit the blood type of his mother?

In the maternity hospital, two children were confused. The first pair of parents has I and II blood groups, the second pair - II and IV. One child has Group II, and the second has Group I. Identify the parents of both children.

Is it possible to transfuse blood to a child from a mother if her blood type is AB and the father's is 00? Explain the answer.

5. Blood group and Rh factor - autosomal unlinked signs. The blood group is controlled by three alleles of one gene - i 0, I A, I B. Alleles IA and IB are dominant with respect to allele i 0. The first group (0) is determined by recessive genes i 0, the second group (A) is determined by the dominant allele IA, the third group (B) is determined by the dominant allele IB, and the fourth (AB) is determined by two dominant allele IAIB. A positive Rh factor dominates over a negative r. The father has a third blood group and is Rh positive (diheterozygote), the mother has a second group and is Rh positive (digomozygote). Determine the genotypes of the parents. What blood group and Rh factor can children in this family have, what are their possible genotypes and the ratio of phenotypes? Make a scheme for solving the problem. What is the law of heredity manifested in this case?

IV

1. In canaries, the presence of a crest is a dominant autosomal trait (A); the sex-linked gene X B determines the green color of the plumage, and X b - brown. In birds, the homogametic sex is male, and the heterogametic sex is female. A crested green female was crossed with a male without a tuft and green plumage (heterozygote). The offspring contained green crested chicks, green without a tufted, brown tufted and brown without a tufted. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, their sex. What laws of heredity are manifested in this case?

2. The body coloration of Drosophila is determined by the autosomal gene. The eye color gene is located on the X chromosome. Male sex is heterogametic in Drosophila. A female with a gray body and red eyes was crossed with a male with a black body and white eyes. All offspring had a gray body and red eyes. The males that turned out in F1 were crossed with the parent female. Make a scheme for solving the problem. Determine the genotypes of the parents and females F 1, the genotypes and phenotypes of the potmy in F 2. What part of females from the total number of offspring in the second crossing is phenotypically similar to the parental female? Indicate their genotypes.

3. In humans, the brown eyes gene dominates over blue eyes (A), and the color blindness gene is recessive (color blindness - d) and linked to the X chromosome. A brown-eyed woman with normal vision, whose father had blue eyes and was color-blind, marries a blue-eyed man with normal vision. Make a scheme for solving the problem. Determine the genotypes of the parents and possible offspring, the likelihood of birth in this family of color-blind children with brown eyes and their gender. Explain the answer.

4. In humans, the inheritance of albinism is not sex-linked (A is the presence of melanin in skin cells, a is albinism), and hemophilia is sex-linked (X n is normal blood clotting, X h is hemophilia). Determine the genotypes of the parents, as well as the possible genotypes, sex and phenotypes of children from a marriage of a diheterozygous woman and an albino man with normal blood clotting for both alleles. Make a scheme for solving the problem. Explain the answer.

1. When crossing maize plants with smooth colored grains with a plant producing wrinkled uncolored seeds, in the first generation, all plants produced smooth colored grains. When analyzing crosses of hybrids from F1 in the offspring there were four phenotypic groups: 1200 smooth colored, 1215 wrinkled uncolored, 309 smooth unpainted, 315 wrinkled colored. Make a scheme for solving the problem. Determine the genotypes of the parents and offspring in the two crosses. Explain the formation of four phenotypic groups in the second cross.

2. When crossing a diheterozygous corn plant with a colored seed and starchy endosperm and a plant with an uncolored seed and a waxy endosperm in the offspring, a phenotype splitting was obtained: 9 plants with colored seed and starchy endosperm; 42 - with colored seed and waxy endosperm; 44 - with uncolored seed and starchy endosperm; 10 - with uncolored seed and waxy endosperm. Make a scheme for solving the problem. Determine the genotypes of the original individuals, the genotypes of the offspring. Explain the formation of the four phenotypic groups.

3. A diheterozygous pea plant with smooth seeds and tendrils was crossed with a plant with wrinkled seeds without tendrils. It is known that both dominant genes (smooth seeds and the presence of antennae) are localized on the same chromosome; crossing over does not occur. Make a scheme for solving the problem. Determine the genotypes of the parents, the phenotypes and genotypes of the offspring, the ratio of individuals with different genotypes and phenotypes. What law is manifested in this case?

VI... Pedigree tasks

1. Based on the pedigree shown in the figure, determine and explain the inheritance pattern of the trait highlighted in black (dominant or recessive, linked or not linked to the sex. Determine the genotypes of offspring 1,3,4,5,6,7. Determine the probability of birth to parents 1 , 2 next children with the trait highlighted in black on the pedigree.

2. Using the pedigree shown in the figure, determine and justify the genotypes of the parents, descendants, indicated on the diagram by the numbers 1,6,7. Establish the probability of having a child with an inherited trait in a woman at number 6, if this trait has never manifested itself in the family of her spouse.

Using the pedigree shown in the figure, determine and explain the inheritance pattern of the trait highlighted in black. Determine the genotypes of parents, offspring1,6 and explain the formation of their genotypes.

Using the pedigree shown in the figure, determine the nature of inheritance of the trait (dominant or recessive, linked or not linked to the sex), highlighted in black, the genotypes of parents and children in the first generation. Indicate which of them is the carrier of the gene, the trait of which is highlighted in black.

Gene interaction problems

1.complementarity

In parrots, feather color is determined by two pairs of genes. The combination of two dominant genes determines the color green. Individuals recessive in both gene pairs are white. The combination of the dominant gene A and the recessive gene b determines the yellow color, and the combination of the recessive gene a with the dominant gene B determines the blue color.

When two green individuals were crossed with each other, they received parrots of all colors. Determine the genotypes of the parents and offspring.

2.epistasis

Rabbit coat color (as opposed to albinism) is determined by the dominant gene. The color of the color is controlled by another gene located in another chromosome, and the gray color dominates over black (in albino rabbits, the color genes do not show themselves).

What are the characteristics of the hybrid forms obtained from crossing a gray rabbit born from an albino rabbit with an albino carrying the gene for black color?

3. pleiotropy.

One of the breeds of chickens is distinguished by short legs. This sign is dominant. The gene that controls it simultaneously causes the shortening of the beak. Moreover, in homozygous chickens, the beak is so small that they are not able to break through the eggshell and die without hatching from the egg. In the incubator of the farm, which breeds only short-legged chickens, 3000 chickens were received. How many are short-legged among them?

4.polymer

The son of a white woman and a black man married a white woman. Could this couple have a child darker than their father?

ANSWERS:

I... Monohybrid crossing problems (complete and incomplete dominance)

1. A - gray color

a - black color P: AA and aa.

F 1: Aa, all gray.

2. A - black floral - white

R: Aa and Aa.

F 1: AA 2Aa, aa. Splitting law.

3. A- white color

a - yellow

Both parents have the Aa genotype.

F 1 genotype cleavage 1aa 2Aa 1 aa

4.Nochka-Aa, her calf-aa. Dawn- aa, her calf- Aa, bull- Aa.

5. A - high

a-dwarf

F 1 1AA 2Aa 1aa

240 dwarf plants.

6.A- green color

a - striped color

F2 1AA 2Aa 1aa

F 2 - cleavage 3 green: 1 striped

7. A- wide leaves

a- narrow leaves

A- intermediate leaf width

F 1: splitting by phenotype: 1- wide leaves, 2- intermediate leaf width, 1- narrow leaves.

By genotype: AA 2Aa aa

II... Problems for dihybrid crossing

1. A - norm, and - glaucoma.
B - Marfan syndrome, b - normal.
One of the spouses suffers from glaucoma and did not have ancestors with Marfan syndrome: aabb. The second spouse is diheterozygous: AaBb.

	norm. syndrome		norm norm	glaucoma syndrome	glaucoma norm

The probability of having a healthy child (norm / norm) \u003d 1/4 (25%). In this case, Mendel's third law is manifested (the law of independent inheritance).

a - dwarfism

B - smooth

in-ribbed
first cross- P: aabb and AaBB, got F 1 - aaBb and AaBb
second - P: aabb and AAbb, got F 1 - Aabb.

4.R-AaBv and aavv.F1: 9 cr. Head .. 3 cr. Op., 3 w., 1 w. op. 4 tons railway op.

5. In this case, intermediate color inheritance appears. AA - black, A - motley, aa - white. parents and hen and rooster have genotypes AaBB. And the gametes form the same: AB, aB. when they merge, the genotypes are formed - AABB - black crested, AaBB - variegated crested, aaBB - white crested. ratio -1/2/1.

color gene:
A - black
a- brown
tail length gene:
B- long
c- short
cc - lethal
BB - shortened
Decision:
1) AaBv x aaBv
black, shortened х brown, shortened
gametes - AB, Av, AV, AV AV AV

AaBB AaBv aaBB aaBv AaBv Aavv aaBv aavv
h. d. h. c. d. c. h lethal c. lethal
black with jlin tail - 1/8
black with shortened - 2/8
brown with a long tail - 1/8
brown with a shortened - 2 /
lethal - 2/8. Independent inheritance law

7. A - mustache

a- beardless

B- red

Bb - pink

1) the first crossing:

R AAVB * aaBB

mustache white b / must. cr.

2) analyzing crossing:

AaBv * aavv

G AV Av av av av

F 2 АаВb - mustachioed pink-fruited; Aabb - whiskered white-fruited;

aaBb - whiskerless pink-fruited; aabb - whiskerless white-fruited;

3) the nature of inheritance of the trait of color of the fetus - incomplete dominance. In the first crossing - the law of uniformity of hybrids, in the second (analyzing) - independent inheritance of traits.

8.A - flat rims,

a - funnel-shaped corollas.

B - white flowers,

b - cream flowers

first cross:

P aaBv x Aavv

G av av av av

F 1 AaBb aavb

second cross:

P aaBB x Aavb

G av av av

F 1 AaBb aaBb
In this case, Medel's third law is manifested - the law of independent inheritance.

9. A - the norm, and - blindness No. 1.

B - norm, b - blindness # 2.
Maternal grandmother is AAbb, paternal grandmother is aaBB. Grandfathers - AABB.

The probability of having a blind grandchild 0%

Tasks for the inheritance of blood groups of the AB0 system

1. Boy-j0j0. Sister- JАJВ

P J A j 0 and J A J B

2. Father - J A J B

Mother-j 0 j 0.

No, because children can have either blood groups 2 or 3.

3.first pair of parents:
P: j 0 j 0 x J А j 0 or j 0 j 0 x J A J А
G: j 0 J A, j 0 j0 J A
F: J A j0 (2 gr.), J 0 j 0 (1 gr.) Or J A j 0 (2 gr.)
second pair of parents
R: J A J A x J A J B or J A j 0 x J A J B
G: J A; J A, J B J A j 0 J A, J B
F: J A J A (2) J A J B (4) J A J A (2) J A J B (4) J A j 0 (2) J in j 0 (3 gr.)
The first couple of parents, the son has 1 gr. and he received gene 0 from both parents. The second pair is the parents of a boy with a 2nd blood group.
This task can be solved orally, because a child with 1 gr. blood cannot be born to a couple in which there is a person with a 4 blood group
4. It is impossible, because children may have blood groups: A0 (II) or B0 (III), therefore, blood of the fourth group, which the mother has, cannot be transfused.

5. Father-diheterozygote I B i 0 Rr, mother-dihomozygote I A I А RR.

	IV group rhesus +	IV group rhesus +	II group rhesus +	II group rhesus +

Children in this family may have blood group IV or II, all Rh-positive. The proportion of children with IV blood group is 2/4 (50%). The law of independent inheritance is manifested (Mendel's third law).

IV... Sex-linked and autosomal inheritance problems

1. A - the presence of a tuft, and - no tuft.
X B - green plumage, X b - brown plumage.
A_X B Y-crested green female
aaX B X b - male without tuft with green plumage (heterozygote)
Among the offspring were chicks without a crest-aa. They got one gene a from their mother, one from their father. Therefore, the mother must have the gene and, therefore, the mother of Aa.
P AaX B Y x aaX B X b

	AaX B X B male	aaX B X B male	AaX B Y female	aaX B Y female
	AaX B X b male	aaX B X b male	AaX b Y female	aaX b Y female

In this case, the law of independent inheritance (Mendel's third law) and sex-linked inheritance appeared.

2. A - gray body

a - black body

X In red eyes

X in - white eyes

P 1 AAX B X B * aaX in Y
gray body black body
red eyes white ch.

G AX B aX in aY
F 1 AaX B X b AaX B Y
P 2 AAX B X B * AaX B Y

G AX B AX B aX B AY aY

F 2 AAX B X B AaX B X B AAX B Y AaX B Y
F 2 all descendants have a gray body and red eyes.

Sex ratio -50%: 50:%

3. A - brown eyes,

a - blue eyes.
X D - normal vision,

X d - color blindness.

A_X D X _ a brown-eyed woman with normal vision
aaX d Y- the woman's father, he could give his daughter only aX d, therefore, the brown-eyed woman is AaX D X d.
AaX D Y. - woman's husband

P AaX D X d x aaX D Y

G AX D AX d aX D ax d aX D aY

F 1 AAX D X D AaX D X d aaX D X D aaX D X d AAX D Y Aa X d Y aaX D XY aaX d Y

The probability of having a color blind child with brown eyes is 1/8, (12.5%), this is a boy.

4. A - the norm, a - albinism.
X N - norm, X h - hemophilia.
Woman AaX H X h, man aaX H Y

G AX H AX h aX H aX h aX H aY

F1 AaX H X H AaX H Y AaX H X h AaX h Y aaX H X H aaX H Y aaX H X h aaX h Y

ph.D. Ph.D. Ph.D. since. Mr. Mr. Mr. yy

Cleavage according to eye color - 1: 1 according to blood clotting - all daughters are healthy, boys - 1: 1.

V. Tasks for chained inheritance

1 . A - smooth grains,

a - wrinkled grains.
B - colored grains,

b - uncolored grains.

RAABB x aavv

Since uniformity was obtained in the first generation (Mendel's first law), therefore, homozygotes were crossed, in F1 a diheterozygote was obtained, bearing dominant characters.

Analyzing cross:

	normal gametes with clutch, a lot
	recombinant gametes with broken clutch, little
		smooth painted, lot (1200)		wrinkled. unpainted., lot (1215)		smooth unpainted., little (309)	wrinkled. painted., little (315)

Since in the second generation, an unequal number of phenotypic groups turned out, therefore, there was linked inheritance. Those phenotypic groups that are represented in large numbers are not crossovers, but groups represented in small numbers are crossovers formed from recombinant gametes, in which the linkage was broken due to crossing over in meiosis.

2. A- colored seed and- uncolored seed B- starchy endosperm b- waxy endosperm P AaBv x aavv G AB Av a B av F1 9- AaBv- env. seed, starch. endosperm 42- Aavv- env. seed, wax. endosperm 44- aaBb- uncolored seed, starchy endosperm 10- aavv- uncolored seed waxy endosperm The presence of two groups in the offspring (42 - with colored waxy endosperm; 44 - with uncolored waxy endosperm) in approximately equal proportions - the result of linked inheritance of alleles , a and B between themselves. Two other phenotypic groups are formed as a result of crossing over.

A - smooth seeds,

a - wrinkled seeds
B - the presence of antennae,

b - without antennae

	smooth seeds, mustache		wrinkled. seeds, no mustache

If crossing over does not occur, then the diheterozygous parent develops only two species of gametes (full linkage).

1. A - gray body

a- black body B- normal wings b- shortened wings P AaBv x aavv

F 1 AaBb are all gray with normal wings. Law of uniformity)

P AaBv x AaBv

T.K. there is no expected Mendelian splitting, which means that crossing over has occurred:

linked

AV AV AV AV
F 2 AABB AaBv AaBv aavv

Genetics, its tasks. Heredity and variability are the properties of organisms. The main genetic concepts... Chromosomal theory of heredity. Genotype as integral system... Development of knowledge about the genotype. The human genome.

Regularities of heredity, their cytological foundations. Mono- and dihybrid crossing. The laws of inheritance established by G. Mendel. Linked inheritance of traits, gene linkage disorder. T. Morgan's laws. Genetics of sex. Inheritance of sex-linked traits. Interaction of genes. Solving genetic problems. Drawing up crossing schemes.

Variability of traits in organisms: modification, mutational, combinative. Types of mutations and their causes. The significance of variability in the life of organisms and in evolution. Reaction rate. The harmful effect of mutagens, alcohol, drugs, nicotine on the genetic apparatus of the cell. Protection of the environment from contamination by mutagens. Identification of sources of mutagens in environment (indirectly) and assessment of the possible consequences of their influence on their own body. Hereditary human diseases, their causes, prevention.

Breeding, its tasks and practical significance. The teachings of N.I. Vavilov on the centers of diversity and origin of cultivated plants. The law of homologous series in hereditary variation. Methods for breeding new varieties of plants, animal breeds, strains of microorganisms. The value of genetics for breeding. Biological bases of cultivation of cultivated plants and domestic animals.

Biotechnology, cell and genetic engineering, cloning. The role of cell theory in the formation and development of biotechnology. The importance of biotechnology for the development of breeding, agriculture, microbiological industry, preservation of the planet's gene pool. Ethical aspects of the development of some research in biotechnology (human cloning, directed genome changes).