Genetics of phenylketonuria. Inheritance

Phenylketonuria

In the liver of healthy people, a small portion of phenylalanine (∼10%) is converted to phenyl lactate and phenylacetylglutamine (Fig. 9-30). This pathway of phenylalanine catabolism becomes the main one when the main pathway is disrupted - the conversion to tyrosine, catalyzed by phenyl-alanine hydroxylase. This disorder is accompanied by hyperphenylalaninemia and an increase in the blood and urine content of alternative pathway metabolites: phenylpyruvate, phenylacetate, phenyllactate and phenylacetylglutamine. A defect in phenylalanine hydroxylase leads to the disease phenylketonuria (PKU). There are 2 forms of PKU:

Classic PKU- a hereditary disease associated with mutations in the phenylalanine hydroxylase gene, which lead to a decrease in the activity of the enzyme or its complete inactivation. In this case, the concentration of phenylalanine increases in the blood by 20-30 times (normally - 1.0-2.0 mg/dl), in the urine - by 100-300 times compared to the norm (30 mg/dl). The concentration of phenylpyruvate and phenyllactate in the urine reaches 300-600 mg/dl with complete absence in the norm.

The most severe manifestations of PKU are impaired mental and physical development, convulsive syndrome, and pigmentation disorders. Without treatment, patients do not live past 30 years. The incidence of the disease is 1:10,000 newborns. The disease is inherited in an autosomal recessive manner.

Severe manifestations of PKU are associated with the toxic effect on brain cells of high concentrations of phenylalanine, phenylpyruvate, and phenyllactate. Large concentrations of phenylalanine limit the transport of tyrosine and tryptophan across the blood-brain barrier and inhibit the synthesis of neurotransmitters (dopamine, norepinephrine, serotonin).

Variant PKU(coenzyme-dependent hyperphenylalaninemia) is a consequence of mutations in genes that control the metabolism of H4 BP. Clinical manifestations are close, but not exactly identical, to those of classical PKU. The incidence of the disease is 1-2 cases per 1 million newborns.

H 4 BP is necessary for the hydroxylation reactions of not only phenylalanine, but also tyrosine and tryptophan, therefore, with a lack of this coenzyme, the metabolism of all 3 amino acids is disrupted, including the synthesis of neurotransmitters. The disease is characterized by severe neurological impairment and early death (“malignant” PKU).

Progressive impairment of mental and physical development in children with PKU can be prevented by a diet that is very low in or eliminates phenylalanine. If such treatment is started immediately after the baby is born, brain damage is prevented. It is believed that dietary restrictions can be eased after the age of 10 (the end of the process of myelination of the brain), but at present many pediatricians are leaning towards a “lifelong diet”.

To diagnose PKU, qualitative and quantitative methods are used to detect pathological metabolites in urine and determine the concentration of phenylalanine in the blood and urine. The defective gene responsible for phenylketonuria can be detected in phenotypically normal heterozygous carriers using a phenylalanine tolerance test. To do this, the subject is given ∼10 g of phenylalanine in the form of a solution on an empty stomach, then blood samples are taken at hourly intervals, in which the tyrosine content is determined. Normally, the concentration of tyrosine in the blood after a phenylalanine load is significantly higher than in heterozygous carriers of the fezylketonuria gene. This test is used in genetic counseling to determine the risk of having an affected child. A screening scheme has been developed to identify newborns with PKU. The sensitivity of the test almost reaches 100%.

Currently, diagnosis of the mutant gene responsible for PKU can be carried out using DNA diagnostic methods (restriction analysis and PCR).

Rice. 9-30. Alternative pathways of phenylalanine catabolism. With a defect in phenylalanine hydroxylase, the accumulated phenylalan and n undergo transamination with α-ketoglutarate. The resulting phenylpyruvate is converted to either phenyllactate or phenylacetylglutamine, which accumulate in the blood and are excreted in the urine. These compounds are toxic to brain cells.

Rice. 9-28. Pathways for the conversion of phenylalanine and tyrosine in different tissues. H 4 BP - tetrahydrobiopterin; H 2 BP - dihydrobiopterin; PF - pyridoxal phosphate; SAM - S-adenosylmethionine.

The genealogical method of studying heredity is one of the oldest and most widely used methods of genetics. The essence of the method is to compile pedigrees that allow one to trace the characteristics of the inheritance of traits. The method is applicable if the direct relatives of the owner of the studied trait on the maternal and paternal lines in a number of generations are known.

Contents 1. 2. 3. 4. 5. Symbols Rules for drawing up a pedigree Stages of problem solving Types of inheritance of characteristics Problem solving

Rules for compiling pedigrees The person from whom they begin to compile a pedigree is called a proband. The proband's brothers and sisters are called sibs. 1. The pedigree is depicted so that each generation is on its own horizontal line. Generations are numbered with Roman numerals, and members of the family tree are numbered with Arabic numerals. 2. Drawing up a pedigree starts from the proband (depending on gender - a square or circle, indicated by an arrow) so that from him it is possible to draw a pedigree both down and up. 3. Next to the proband, place the symbols of his siblings in order of birth (from left to right), connecting them with a graphic rocker.

4. Above the proband line, indicate the parents, connecting them to each other with a marriage line. 5. On the parents’ line, draw the symbols of the closest relatives and their spouses, connecting their degrees of relationship accordingly. 6. On the proband’s line, indicate his cousins, etc., brothers and sisters, connecting them accordingly with the parents’ line. 7. Above the line of parents, draw the line of grandparents. 8. If the proband has children or nephews, place them on a line below the proband's line.

9. After depicting the pedigree (or simultaneously with it), appropriately show the owners or heterozygous carriers of the trait (most often, heterozygous carriers are determined after the compilation and analysis of the pedigree). 10. Indicate (if possible) the genotypes of all members of the pedigree. 11. If there are several hereditary diseases in the family that are not related to each other, create a pedigree for each disease separately.

Stages of problem solving 1. Determine the type of inheritance of the trait - dominant or recessive. To do this, find out: 1) whether the trait being studied is common (in all generations or not); 2) how many members of the pedigree have the trait; 3) whether there are cases of birth of children possessing the trait, if the parents do not exhibit this trait; 4) whether there are cases of birth of children without the studied trait, if both parents have it; 5) what part of the offspring carries the trait in families if one of the parents is its owner.

Stages of problem solving 2. Determine whether the trait is inherited in a sex-linked manner. To do this, find out: 1) how often the symptom occurs in people of both sexes; if it is rare, then which gender carries it more often; 2) persons of which gender inherit the trait from the father and mother who carry the trait.

Stages of problem solving 3. Based on the results of the analysis, try to determine the genotypes of all members of the pedigree. To determine genotypes, first of all, find out the formula for the splitting of descendants in one generation.

Types of inheritance of a trait. 1. Autosomal dominant inheritance: 1) the trait occurs frequently in the pedigree, in almost all generations, equally often in both boys and girls; 2) if one of the parents is a carrier of a trait, then this trait will appear either in all of the offspring or in half.

Glaucoma is an eye disease characterized by increased intraocular pressure and decreased visual acuity. Risk factors for the development of glaucoma are: heredity, diabetes mellitus, atherosclerosis, eye trauma, inflammatory and degenerative eye diseases. With constantly elevated intraocular pressure, atrophy of the optic nerve gradually develops, and the person loses vision. Brachydactyly (brachydactylia; brachy- + Greek daktylos finger; synonym short-fingered) is a developmental anomaly: shortening of the fingers or toes. inherited in an autosomal dominant manner.

Types of inheritance of a trait. 2. Autosomal recessive inheritance: 1) the trait is rare, not in all generations, equally common in both boys and girls; 2) the trait can appear in children, even if the parents do not have this trait; 3) if one of the parents is a carrier of the trait, then it will not appear in children or will appear in half of the offspring.

What is phenylketonuria? Phenylketonuria (PKU) is an inherited disorder that increases the amount of the amino acid phenylalanine in the blood to harmful levels. (Amino acids are the building blocks of proteins). If PKU is not treated, excess phenylalanine can cause mental retardation and other serious health problems. How do people inherit PKU? PKU is inherited in an autosomal recessive manner, which means two copies of the gene must be changed for a person to be affected by the disease. Most often, the parents of a child with an autosomal recessive disorder are not affected, but are carriers of one copy of the altered gene.

Types of inheritance of a trait. 3. Sex-linked inheritance: 1) X - dominant inheritance: ü the trait is more common in females; ü if the mother is sick and the father is healthy, then the trait is transmitted to the offspring regardless of gender; it can manifest itself in both girls and boys; ü if the mother is healthy and the father is sick, then all daughters will exhibit the symptom, but sons will not.

3. Sex-linked inheritance: 2) X - recessive inheritance: the trait is more often found in males; More often the symptom manifests itself after a generation; If both parents are healthy, but the mother is heterozygous, then the trait often appears in 50% of sons; If the father is sick and the mother is heterozygous, then female persons can also have the trait.

3. Sex-linked inheritance: 3) Y-linked inheritance: ütrait occurs only in males; If the father carries a trait, then, as a rule, all sons also possess this trait.

An example of solving the problem The proband is a right-handed woman. Her two sisters are right-handed, her two brothers are left-handed. Mother is right-handed. She has two brothers and a sister, all right-handed. Grandmother and grandfather are right-handed. The proband's father is left-handed, his sister and brother are left-handed, the other two brothers and sister are right-handed. Solution: 1. Draw the symbol of the proband. We show the presence of the sign in the proband.

2. We place the symbols of her siblings next to the proband symbol. We connect them with a graphic rocker.

7. Determine the genotypes of the pedigree members. The sign of right-handedness appears in every generation in both females and males. This indicates an autosomal dominant type of inheritance of the trait. I A- A- II A- A- A- Aa aa A- III aa Aa Aa A- aa

Task 2. Based on the pedigree shown in the figure, determine the nature of the manifestation of the trait indicated in black (dominant, recessive, sex-linked or not). Determine the genotype of parents and children in the first generation.

Scheme for solving the problem: 1) The recessive trait is not sex-linked; 2) Genotypes of the parents: mother - aa, father - AA or Aa 3) Genotypes of the children: heterozygous son and daughter - Aa.

Task 3 Using the pedigree shown in the diagram, establish the type and nature of manifestation of the trait highlighted in black (dominant, recessive, sex-linked or not). Determine the genotypes of children in the first generation.

Scheme for solving the problem: 1) The trait is recessive, linked to the X chromosome; 2) Genotypes of the parents: mother – XHA, father – XAU; 3) Genotypes of children in F 1: son - Ha. Uh, daughter - HAHA daughter - HAHA

Task 4 Using the person’s pedigree shown in the figure, establish the nature of inheritance of the “small eyes” trait, highlighted in black (dominant or recessive, sex-linked or not). Determine the genotypes of parents and offspring F 1 (1, 2, 3, 4, 5). 1 2 3 4 5

Scheme for solving the problem: 1) The trait is recessive, not sex-linked; 2) Genotypes of the parents: mother – Aa, father – Aa; 3) Genotypes of descendants in F 1: 1, 2 – Aa, 3, 5 – AA or Aa; 4 – aa.

Codifier of content elements in biology 3. 4 Genetics, its tasks. Heredity and variability are properties of organisms. Genetics methods. Basic genetic concepts and symbolism. Chromosomal theory of heredity. Modern ideas about the gene and genome. 3. 5 Patterns of heredity, their cytological basis. Patterns of inheritance established by G. Mendel, their cytological basis (mono- and dihybrid crossing). Morgan's laws: linked inheritance of traits, disruption of gene linkage. Genetics of sex. Inheritance of sex-linked traits. Gene interaction. Genotype as an integral system. Human genetics. Methods for studying human genetics. Solving genetic problems. Drawing up crossing schemes.

SPECIFICATION of examination paper in biology A 7. Genetics, its tasks, basic genetic concepts. A 8. Patterns of heredity. Human genetics. A 9. Patterns of variability. A 30. Genetic patterns. The influence of mutagens on the genetic apparatus of cells and organisms. C 6. Solving problems in genetics to apply knowledge in a new situation.

Part A 1. Genetics is of great importance for medicine, since it 1) fights epidemics 2) creates medicines to treat patients 3) establishes the causes of hereditary diseases 4) protects the environment from pollution by mutagens

2. The method used to study the nature of the manifestation of characteristics in sisters or brothers who developed from one fertilized egg is called 1. 2. 3. 4. Hybridological Genealogical Cytogenetic Twin

3. The genealogical method is used for 1) Obtaining gene and genomic mutations 2) Studying the influence of education on human ontogenesis 3) Researching hereditary human diseases 4) Studying the stages of evolution of the organic world

4. What is the function of medical genetic consultations for parental couples? 1. Identifies the predisposition of parents to infectious diseases 2. Determines the possibility of having twins 3. Determines the likelihood of hereditary diseases in children 4. Identifies the predisposition of parents to metabolic disorders

Determine genotype by phenotype Eye color in a person is determined by an autosomal gene; color blindness is a recessive gene linked to sex. Determine the genotype of a brown-eyed woman with normal color vision, whose father is color-blind (brown-eyedness dominates blue-eyedness) 1) AAXDXD 3) Aa. Xd 2) Aa. XDXd 4) aa. XDXd

Part C Solution of genetic problems on the application of knowledge in a new situation: dihybrid crossing, inheritance of sex-linked traits, linked inheritance of traits (with crossing over, without crossing over), determination of blood groups, pedigree analysis

Part C In humans, the inheritance of albinism is not sex-linked (A - the presence of melanin in skin cells, and - the absence of melanin in skin cells - albinism), and hemophilia is sex-linked (XH - normal blood clotting, Xh - hemophilia). Determine the genotypes of the parents, as well as the possible genotypes, sex and phenotypes of children from the marriage of a dihomozygous woman, normal for both alleles, and an albino man with hemophilia. Make a diagram for solving the problem.

The scheme for solving the problem includes: 1) genotypes of the parents: ♀AAXHXH (AXH gametes); ♂aa. Xh. Y (gametes a. Xh, a. Y); 2) genotypes and gender of children: ♀Aa. XHXh; ♂Aa. XHY; 3) phenotypes of children: a girl who is outwardly normal for both alleles, but is a carrier of the genes for albinism and hemophilia; A boy who is outwardly normal for both alleles, but is a carrier of the albinism gene.

The law of cleavage also explains the inheritance of phenylketonuria
(PKU) - a disease that develops as a result of an excess of an important
amino acids - phenylalanine (Phe) in the human body. Excess
phenylalanine leads to the development of mental retardation. Frequency
The incidence of PKU is relatively low (approximately 1 in 10,000 new
born), however, about 1% of mentally retarded individuals
mov suffer from PKU, thus making up a relatively more
largest group of patients whose mental retardation is explained
homogeneous genetic mechanism.
As in the case of CG, researchers studied the frequency of occurrence
PKU in families of probands. It turned out that patients suffering from PKU
usually have healthy parents. In addition, it was noticed that
PKU is more common in families in which parents are blood
other relatives. Example of a family of a proband suffering from PKU

73 is shown in Fig. 2.3: sick
the child was born with a phenotypic
healthy parents -
blood relatives (two-
cousins), but
the child's father's sister is suffering
PKU.
PKU is transmitted recessively
strong type of inheritance,
those. the patient's genotype contains
two PKU alleles obtained
from both parents. Descendants,
who have only one
such an allele do not suffer from
disease, but are but-
carriers of the PKU allele and may
pass it on to your children. On
rice. 2.4 shows the paths of inheritance
formation of PKU alleles from two
phenotypically normal
parents. Each of the parents
leu has one PKU allele and one normal allele. Probability
that every child can inherit the PKU allele from every
of the parents is 50%. The probability that the child is
follows the PKU allele from both parents at the same time, is 25%
(0.5 x 0.5 = 0.25; probabilities are multiplied as events are inherited
the alleles from each parent are independent of each other).
The PKU gene and its structural variants found in different
populations have been well studied. The knowledge at our disposal is

Rice. 2.4. Crossing scheme: allelic mechanism of inheritance of PKU.
F - dominant allele (“healthy”); [f] - recessive allele causing
the development of the disease. FF, FF - phenotypically normal children (75% of them); only
about 25% have a normal genotype (FF); another 50% are phenotypically healthy,
but are carriers of the PKU (FF) allele. The remaining 25% of descendants are sick
([f][f]).

74
Rice. 2.3. An example of a family pedigree, in
which PKU is transmitted via
inheritance (the proband's aunt suffers
this disease).
A double line between spouses means
consanguineous marriage. Rest
the designations are the same as in Fig. 2.1. marriage, allow for timely prenatal diagnosis
tics in order to determine whether the developing embryo has inherited
breathe two copies of the PKU allele from both parents (the fact of such inheritance
vaniya sharply increases the likelihood of disease). In some countries,
for example in Italy, where the incidence of PKU is quite high
juice, such diagnostics are carried out without fail for each
milk a pregnant woman.
As already noted, PKU is more common among those who enter
marries with blood relatives. Despite the fact that the meeting
The incidence of PKU is relatively low, approximately 1 in 50 people is
carrier of the PKU allele. The probability that one carrier of the allele
PKU will marry another carrier of such an allele, is
approximately 2%. However, when marrying between consanguineous
relatives (i.e. if the spouses belong to the same pedigree, in
which PKU allele is inherited) the probability that
both spouses will be carriers of the PKU allele and at the same time transfer
will give two alleles to the unborn child, it will become significantly higher than 2%.

You look at the article (abstract): “ RECESSIVE INHERITANCE: PHENYLKETONURIA"from discipline" Psychogenetics»

Abstracts and publications on other topics :

A number of gene mutations, in which the structure of only one gene is changed, leads to the development of mental retardation. According to some estimates, in 7-10% of patients with oligophrenia it is caused by mutations of this kind.

The set of biochemical reactions occurring in the body is called metabolism. Many genes encode proteins that participate as enzymes in certain metabolic reactions. A mutation in such a gene can lead to the body producing a less active or completely inactive enzyme, and sometimes to a complete cessation of enzyme synthesis. In this case, the reaction, normally carried out by this enzyme, either slows down or does not occur at all, which causes the corresponding hereditary disorder - one of the so-called inborn errors of metabolism. The most common genetic hereditary diseases include phenylketonuria, sickle cell anemia, Tay-Sachs disease, hemophilia, and diabetes mellitus. The extent to which they influence the phenotype depends on how important the affected enzyme is to the organism. We saw above that Tay-Sachs disease and cystic fibrosis lead to death. Some other genetic abnormalities cause various serious problems in the body, but are not fatal.

Phenylketonuria and albinism affect the same metabolic pathway.

Phenylketonuria is a disease in which, as a result of a mutation, the structure of the enzyme involved in the metabolism of the amino acid phenylalanine (phenylalanine hydroxylase) is disrupted. This enzyme is necessary for the conversion of phenylalanine to tyrosine. Diseases of this kind are called enzymopathies, i.e. caused by a defect in enzymes. With this disease, phenylalanine and products of its improper metabolism (phenylacetic acid) accumulate in the blood, which leads to damage to the developing nervous system. This is mainly the destruction of myelin and degeneration of the spongiform nervous system. Mental retardation, microcephaly, psychosis, tremor, convulsive activity, and spasticity occur.

Phenylketonuria affects individuals who are homozygous for a recessive gene that deprives them of the ability to synthesize one of the enzymes necessary to convert the amino acid phenylalanine into another amino acid, tyrosine. Instead of being converted to tyrosine, phenylalanine is converted to phenylpyruvic acid, which accumulates in toxic quantities in the blood, affects the brain and (if not treated promptly) causes mental retardation. The urine of patients also contains phenylpyruvic acid, which gives it a characteristic odor. Currently, phenylketonuria is treated with a special diet. To do this, in the first years of a child’s life, phenylalanine is almost completely excluded from his diet. Once brain development is complete, a patient with phenylketonuria is placed on a normal diet, but a woman with this genetic disorder should follow a diet low in phenylalanine during pregnancy to prevent abnormal development of the fetal brain. In the United States, in many states, all newborns are required to undergo special tests for PKU and some other inborn errors of metabolism.

Individuals homozygous for the albinism gene lack the enzyme that normally catalyzes the conversion of tyrosine to melanin, i.e. pigment that determines the brown or black color of eyes, hair and skin. Albinos have white hair and very light skin and eyes. Naturally, the question may arise whether patients with phenylketonuria are also albinos, since their bodies do not produce tyrosine, from which melanin is ultimately produced. However, such patients are not albinos, because tyrosine is not only formed in the body itself from phenylalanine, but also enters the body with food. True, patients with phenylketonuria are usually light-eyed, fair-skinned and fair-haired. Of course, there may be albinos among them, but only if the individual is homozygous for both recessive genes.

Inheritance of phenylketonuria (PKU) explains the law of splitting. This mutation is recessive, i.e. can resist in phenotype only in the homozygous state. The highest incidence of phenylketonuria was observed in Ireland (16.4 cases per 100 thousand newborns); for comparison: in the USA - 5 cases per 100 thousand newborns.

The PKU gene and its structural variants, found in different populations, have been well studied. The knowledge at our disposal allows us to carry out timely pre-natal diagnosis in order to determine whether the developing embryo has inherited two copies of the PKU allele from both parents (the fact of such inheritance sharply increases the likelihood of the disease). In some countries, for example in Italy, where the incidence of PKU is quite high, such diagnosis is mandatory for every pregnant woman.

PKU is more common among those who marry blood relatives. Although the incidence of PKU is relatively low, approximately 1 in 50 people are carriers of the PKU allele. The probability that one carrier of the PKU allele will marry another carrier of such an allele is approximately 2%. However, when a marriage occurs between blood relatives (i.e., if the spouses belong to the same pedigree in which the PKU allele is inherited), the likelihood that both spouses will be carriers of the PKU allele and simultaneously pass on two alleles to the future child will become significantly higher 2 %.

In the case of phenylketonuria, we have a striking example of how the development of a disease that has a genetic nature can be prevented by selecting environmental influences. Currently, phenylketonuria is easily detected during routine examinations of newborns at 2-3 days of age (normally, the concentration of phenylalanine in the blood plasma should not exceed 4 mg/dL). Patients are placed on a diet low in phenylalanine, which helps avoid developmental damage to the nervous system. In this case, tyrosine becomes an essential amino acid and it is necessary to ensure its presence in the diet. The most critical period is the early stages of ontogenesis, therefore, in adulthood, many no longer adhere to dietary restrictions, although this is still desirable. Women with phenylketonuria, regardless of their own condition, must follow a special diet during pregnancy, otherwise the high levels of phenylalanine in their blood will have a damaging effect on the developing fetus.

Phenylketonuria is a good example of a genotype-environment interaction. The essence of this disease is the different sensitivity of individuals with different genotypes to environmental influences. The same environment (in this case, the environment is the nature of nutrition) causes a severe illness (phenylketonuria) in some genotypes, while in other genotypes absolutely no pathological changes are observed. Under other environmental conditions (subject to a special diet), the differences between genotypes for this trait (phenylketonuria) disappear.

1. Consider the features of studying human genetics, develop knowledge about the basic methods of studying human heredity.
2. Develop the ability to use genetic terms and patterns to solve practical problems.
3. Deepening and expanding students' knowledge of medical genetics, awareness of the importance of acquired knowledge for the health of future generations.

• explanatory and illustrative (conversation, story, demonstration of notes and genealogy diagrams);
• research (solving problems, analyzing pedigree charts, drawing up pedigree charts);
• group work.

Equipment:

• collection “Problems in Genetics”;
• student information sheets “Human Genetics”;
• pedigree charts;
• table “Inheritance of the hemophilia gene in the royal houses of Europe.”

Lesson type: Lesson on applying knowledge and skills.

During the classes

I. Updating knowledge

Teacher’s opening statement: “For genetic research, a person is a very inconvenient object: a large number of chromosomes, experimental crossing is impossible, puberty comes late, a small number of descendants in each family. Today at the lesson there are representatives of four independent genetic laboratories who will help us figure out the basic methods for studying human heredity.”

One student per group is invited to the board to solve problems.

Task 1. Phenylketonuria (a disorder of phenylalanine metabolism, which results in dementia) is inherited as an autosomal recessive trait. What can children be like in a family where parents are heterozygous for this trait?

Answer: The probability of having healthy children is 75%, sick - 25%.

Task 2. Sickle cell anemia (a change from normal hemoglobin - A to S-hemoglobin, causing red blood cells to take the shape of a sickle) is inherited as an incompletely dominant autosomal gene. The disease in homozygous individuals leads to death, usually before puberty; heterozygous individuals are viable, and their anemia most often manifests itself subclinically. Interestingly, the malarial plasmodium cannot use S-hemoglobin for its nutrition. Therefore, people who have this form of hemoglobin do not get malaria.

Answer: The probability of having children who are not resistant to malaria is 25%.

Task 3. Classic hemophilia is transmitted as a recessive, X-linked trait. A healthy man marries a woman whose brother has hemophilia. Determine the probability of having healthy children in this family.

Answer: The probability of having healthy children is 75%.

Task 4. Enamel hypoplasia (thin, granular enamel, light brown teeth) is inherited as an X-linked dominant trait. In a family where both parents suffered from this anomaly, a son was born with normal teeth. Determine the probability that their next child will also have normal teeth.

Answer: The probability of having healthy children is 25%.

II. Methods for studying human genetics

Representatives of genetic laboratories explain the essence of genetic methods using the information sheet “Human Genetics” and the table “Inheritance of the hemophilia gene in the royal houses of Europe.”

1. Twin method

Twins are children born at the same time. They are monozygotic (identical) and dizygotic (fraternal). Monozygotic twins develop from one zygote, which at the cleavage stage is divided into two (or more) parts. Therefore, such twins are genetically identical and always of the same sex. Monozygotic twins are characterized by a high degree of similarity (concordance) for many characteristics. The degree of concordance for qualitative traits in monozygotic twins is usually high and tends to 100%. This means that the environment has almost no influence on the formation of characteristics of blood groups, eyebrow shape, eye and hair color, and the genotype has a decisive influence. The twin method confirmed the hereditary causes of hemophilia, diabetes mellitus, and schizophrenia. A pronounced predisposition to a number of diseases has been discovered: tuberculosis, rheumatism, etc., which means a greater likelihood of these diseases occurring in people with a certain genotype under favorable conditions

Dizygotic twins develop from eggs that were simultaneously ovulated and fertilized by different sperm. Therefore, they are hereditarily different and can be either the same or different sexes. They are dissimilar (discordant) in many ways.

Observations of twins provide material for identifying the role of heredity and environment in the development of traits.

2. Genealogical method

The essence of the method is to study pedigrees in those families in which there are hereditary diseases. The method allows you to determine the type of inheritance of a trait and, based on the information obtained, predict the likelihood of the manifestation of the studied trait in the offspring, which is of great importance for the prevention of hereditary diseases.

Thanks to a well-known pedigree, it was possible to trace the inheritance of the hemophilia gene from Queen Victoria of England. Victoria and her husband were healthy. It is also known that none of her ancestors suffered from hemophilia. Most likely, a mutation occurred in the gamete of one of Victoria’s parents. As a result, she became a carrier of the hemophilia gene and passed it on to many descendants. All male descendants who received an X chromosome with a mutant gene from Victoria suffered from hemophilia. The hemophilia gene is recessive and is inherited linked to the X chromosome.

The following diseases are inherited in an autosomal dominant manner: glaucoma, achondroplasia, polydactyly (extra fingers), brachydactyly (short fingers), arachnodactyly (Morphan's syndrome).

According to the autosomal recessive type inherited: albinism, phenylketonuria, allergies, schizophrenia.

X-linked dominant traits: enamel hypoplasia (thin granular enamel, light brown teeth).

X-linked recessive traits: hemophilia, color blindness, lack of sweat glands.

Y-linked features: hypertrichosis(hair growth on the edge of the auricle), syndactyly (fusion of the fingers).

The use of the genealogical method has shown that with related marriages, compared with unrelated ones, the likelihood of deformities, stillbirths, and early mortality in the offspring significantly increases, since recessive genes more often become homozygous.

3. Cytogenetic method

Based on the study of the human chromosome set. Normally, a human karyotype includes 46 chromosomes - 22 pairs of autosomes and two sex chromosomes. The use of this method made it possible to identify a group of diseases associated with either a change in the number of chromosomes or a change in their structure.

Patients with Klinefelter syndrome(47, XXY) always men. They are characterized by underdevelopment of the gonads, often mental retardation, and high growth (due to disproportionately long legs).

Shereshevsky-Turner syndrome(45, XO) is observed in women. It manifests itself in delayed puberty, underdevelopment of the gonads, absence of menstruation, and infertility. Women with Shereshevsky-Turner syndrome are small in stature, have broad shoulders, a narrow pelvis, shortened lower limbs, a short neck with folds, and a “Mongoloid” eye shape.

Down syndrome- one of the most common chromosomal diseases. It develops as a result of trisomy on chromosome 21 (47,21,21,21); the disease is easily diagnosed, as it has a number of characteristic signs: shortened limbs, small skull, flat, wide bridge of the nose, narrow palpebral fissures with an oblique incision, the presence of a fold of the upper eyelid , mental retardation.

Most often, chromosomal diseases are the result of mutations that occur in the germ cells of one of the parents during meiosis.

4. Biochemical method

The method involves determining the activity of enzymes or the content of certain metabolic products in the blood or urine. Using this method, metabolic disorders that occur in various pathological conditions and are caused by the presence of an unfavorable combination of allelic genes in the genotype are identified. For example, the severe disease phenylketonuria develops in the case of homozygosity for a recessive gene, the activity of which blocks the conversion of the essential amino acid phenylalanine into tyrosine, and phenylalanine is converted into phenylpyruvic acid, which is excreted in the urine. The disease leads to the rapid development of dementia in children. Early diagnosis and a phenylalanine-free diet can slow the progression of the disease. In heterozygotes for the phenylketonuria gene, an increased content of phenylalanine is found in the blood, although the phenotype does not change significantly, the person is healthy. In hemophilia, heterozygous carriage of a mutant gene can be established by determining the activity of the enzyme altered as a result of the mutation. Thus, using the biochemical method, it is possible to predict with great accuracy the risk of offspring with this disease.

• Checking the problems solved on the board.

III. Pedigree analysis

Students receive an assignment.

1. Learn the principles of genealogical analysis

In human genetics, an analogue of the hybridological method is genealogical analysis. It consists of compiling and studying a graphical representation of pedigrees, each of which reflects family ties between healthy and sick people of several generations. Males are indicated by squares, and females are indicated by circles. People who have the studied trait (for example, a disease) in their phenotype are depicted as black figures, and those who have an alternative trait are white. Some horizontal lines connect graphic images of spouses with each other, and others - graphic images of their children. Vertical lines connect graphic images of parents and their children to each other.

2. Study family pedigree graphics according to one studied characteristic. A member of this family who applied for medical genetic consultation and is called a proband is indicated in the diagram by an arrow.

Pedigree 1

Pedigree 2

3. Answer the following questions

1. How many generations of people are represented in the graphical representation of the proband’s pedigree?

2. How many children did the proband’s paternal grandparents have?

3. What is the gender of the proband?

• 1-male
• 2-female

4. Does the proband have the studied trait?

• 1 - yes
• 2 - no

5. How many other members of the pedigree have the same trait as the proband?

6. Is the trait being studied recessive or dominant?

• 1 - recessive
• 2 - dominant

7. Name the chromosome that contains the allele responsible for the formation of the trait being studied

• 1 - autosome
• 2 - X chromosome
• 3 -Y chromosome

8. What is the genotype of a) the proband, b) the proband’s brother, c) the proband’s mother, d) the proband’s father?

1-AA	5-X A X A	9-XX
2-Aa	6-X a X a	10-XY a
3-aa	7-X A Y	11-XY
4-X A X a	8-X a Y

4. Discuss the answer in groups

Pedigree 1: 1- 3; 2-5; 3-2; 4-1; 5-8; 6-2; 7-1; 8-a) 2 b) 3 c) 3 d) 2;

Pedigree 2: 1- 4; 2-6; 3-1; 4-1; 5-6; 6-1; 7-2; 8-a) 8 b) 7.8 c) 4 d) 7;

5. Make a graphic representation of the pedigree

Rosa and Alla are sisters and both, like their parents, suffer from night blindness. They also have a sister with normal vision, and a brother and sister who suffer from night blindness. Rosa and Alla married men with normal vision. Alla had two girls and four boys suffering from night blindness. Rosa has two sons and a daughter with normal vision and another son suffering from night blindness.

Determine the genotypes of Rose, Alla, their parents and all children.

Answer: Father's genotype - Aa; mother - Aa; Rose - Aa; Alla - AA; the sister of Rosa and Alla, who does not suffer from night blindness, has the genotype - aa; another sister and brother - AA or Aa; all Alla's children are Aa; Rose's children who do not suffer from night blindness - aa, son - Aa.

Graphic representation of the pedigree of Rose and Alla.

IV. Summing up the lesson

Human genetics is one of the most rapidly developing branches of science. It is the theoretical basis of medicine and reveals the biological basis of hereditary diseases. Knowledge of the genetic nature of diseases makes it possible to make an accurate diagnosis in time and carry out the necessary treatment, preventing the birth of sick children. In the next lesson we will talk about social problems of human genetics. Now let's summarize the lesson. Please hand in your score sheet.

V. Homework

P. 35, creative works on human genetics (abstracts, messages, newspapers, bulletins, video clips).

Application

Literature

1. Biology: Textbook for medical students. school/ V.N. Yarygin-M. Humanit. ed. VLADOS center, 2001.
2. We are preparing for the Unified State Exam. General biology./V.N. Frosin, V.I. Sivoglazov. -M.: Bustard. 2004.
3. Problem book on general and medical genetics./N. V. Helevin, -M.: Higher School, 1976.
4. General biology. Textbook for 10-11 grades of secondary school / D.K. Belyaev, G.M. Dymshits, - M.: Education, 2005.
5. General biology. A textbook for grades 10-11 with in-depth study of biology at school./ V.K. Noisy, G. M. Dymshits, A. O. Ruvinsky, - M.: Education, 2001.