Hip dysplasia: how to identify and can it be cured?

Congenital hip dislocation - symptoms and treatment

Congenital hip dislocation is a pathology of the hip joints associated with their congenital atypical structure. The disease begins to develop from the first days of life and is usually diagnosed in childhood. Congenital hip dislocation is characterized by scant manifestations at the very beginning, steady progression and complete destruction of the hip joints in the absence of timely treatment [1].

A joint is a movable joint between two bones; its normal operation is possible only if all its components are in full compliance with each other. In the hip joint, the head of the femur must be shaped to fit the articular surface of the pelvis (acetabulum). The separation of the “hinge” of connecting segments (articular surfaces of bones) is called a dislocation. [4]. When a sudden dislocation occurs as a result of injury, the soft tissue around the joint is stretched, and ligaments, muscles and tendons are torn under the skin.

Congenital hip dislocation is a similar process, only it proceeds slowly. Congenital dislocation is in no way connected with injury: for its development, ordinary household activities are sufficient: walking, running, jumping and active play of the child. In this case, the joint becomes vulnerable due to disrupted anatomy [1].

Changes in the joint elements are called hip dysplasia We are talking about dysplasia when:

• the depth of the glenoid cavity is too small;
• the edges of the glenoid cavity are excessively sloping;
• the ratio of joint angles is disturbed;
• The cartilaginous elements that stabilize the joint do not sufficiently perform their stabilizing function.

Experts sometimes call such joints “ dysplastic . Such a joint develops too slowly, it is “softer” than a healthy joint characteristic of this age. The characteristics of the second part of the joint, the articular head, also do not correspond to the norm. In this case, it is not hard enough, has an atypical, oval shape, does not tolerate loads well and accepts them only partially, and not evenly over the entire surface.

All of the above causes instability of the joint: such anatomy of the hip joint is not able to withstand the loads intended for it [20]. Dysplasia of the hip joint creates a favorable background and inevitable (in the absence of treatment) flow into dislocation of the hip joint [6]. The articulating surfaces are not held in the desired position and gradually become separated as the child grows, and the dislocation progresses. This happens especially quickly during critical periods: when the child begins to sit down, stand up and walk, creating a load on the joints.

The main reason for the formation of congenital hip dislocation is a violation of the correct formation of the joint (dysplasia), as a result of which the anatomy and biomechanics of the joint change. The exact causes of failures in the formation of the hip joints are unknown. Versions have been put forward about intrauterine developmental disorders associated with unfavorable factors during pregnancy:

• smoking, drinking alcohol or using drugs;
• viral diseases;
• malnutrition;
• obstetric infections;
• breech presentation of the fetus.

Signs and symptoms of hip dysplasia

Symptoms of severe degrees of dysplasia (subluxation and dislocation) are pronounced in children. To exclude the diagnosis, the child should be examined by a pediatric orthopedist for preventive purposes. The “golden” standard for making a diagnosis is the maternity hospital! In the first year of life, an orthopedic examination is carried out at 1, 3, 6 and 12 months.

Main specific clinical signs in a child 1-2 months old

• Barlow test and Ortolani test

Additional (less specific) clinical signs

• Asymmetrical arrangement of skin folds on the thighs
• Asymmetrical location of the gluteal folds
• Excessive hip rotation
• External rotation of feet
• Oblique location of the genital fissure in girls.
• Shortening of the lower limb
• Palpation of the femoral head behind the posterior edge of the socket
• Limitation of abduction of legs bent at right angles at the hip and knee joints

Diagnostics

In a baby, signs of hip dysplasia in the form of a dislocation can be diagnosed in the maternity hospital. The neonatologist should carefully examine the child for the presence of such abnormalities in certain pregnancy complications.

The risk group includes children who belong to the category of large children, children with deformed feet and those with heredity burdened by this characteristic. In addition, attention is paid to toxicosis of pregnancy in the mother and the gender of the child. Newborn girls are subject to mandatory examination.

Examination methods:

1. Ultrasound diagnostics is an effective method for identifying abnormalities in the structure of joints in children in the first three months of life. Ultrasound can be performed multiple times and is acceptable when examining newborns. The specialist pays attention to the condition of the cartilage, bones, joints, and calculates the angle of the hip joint.
2. Arthroscopy and arthrography are performed in severe, advanced cases of dysplasia. These invasive techniques require general anesthesia to obtain detailed information about the joint.
3. CT and MRI provide a complete picture of pathological changes in the joints in various projections. The need for such an examination appears when planning surgical intervention.
4. The X-ray image is not inferior in reliability to ultrasound diagnostics, but has a number of significant limitations. The hip joint in children under seven months of age is poorly visible due to the low level of ossification of these tissues. Radiation is not recommended for children in their first year of life. In addition, placing an active baby under the device while maintaining symmetry is problematic.
5. External examination and palpation are carried out to identify characteristic symptoms of the disease. In infants, hip dysplasia has signs of both dislocation and subluxation, which are difficult to identify clinically. Any symptoms of abnormalities require a more detailed instrumental examination.

Prevention of pathology

If you do not want dysplasia to appear in your baby, you must take certain precautions:

1. Taking vitamins, proper nutrition, light physical activity during pregnancy.
2. Constantly following your doctor's recommendations during pregnancy. In this case, an important element of the examination is ultrasound, which can show health problems at an early stage of fetal development.
3. Postpartum examination by an orthopedist, as well as an ultrasound of the hip joint.
4. It is necessary to eliminate the causes that can lead to the appearance of pathology and provoke dislocation.
5. The use of therapeutic exercises and regular physical activity, which will help place and fix the bone in place.
6. Carrying a baby in a sling, as well as using wide swaddling.
7. If the diagnosis of “dysplasia” is nevertheless made, then the baby cannot be put on his feet until the doctor allows it.

Modern methods of diagnosing and treating hip dysplasia are still far from perfect. In outpatient settings (clinics), cases of underdiagnosis (the diagnosis is not made in time for existing pathology) and overdiagnosis (the diagnosis is made in healthy children) are still common.

Many orthopedic structures and surgical treatment options have been proposed. But none of them can be called completely perfect. There is always a certain risk of relapses and complications. Different clinics practice different approaches to the diagnosis and treatment of pathology. Currently, research continues to be actively conducted.

Severity of traffic accidents

1. I degree – pre-dislocation. A developmental deviation in which the muscles and ligaments are not changed, the head is located inside the beveled cavity of the joint.
2. II degree – subluxation. Only part of the femoral head is located inside the articulation cavity, as it moves upward. The ligaments are stretched and lose tension.
3. III degree – dislocation. The head of the femur comes completely out of the socket and is located higher. The ligaments are tense and stretched, and the cartilaginous rim fits inside the joint.

Physiotherapy

Many physiotherapeutic procedures are used that eliminate the inflammatory reaction, improve joint trophism and reduce joint pain. The most commonly used procedures are:

Wearing various orthopedic devices

Freik's pillow, Pavlik's stirrups and others. All this also helps to keep the baby's legs spread and bent. It is this method of treating hip dysplasia in infants that seems blasphemous to many parents, since they have to constantly see their baby “shackled” in orthopedic spacers.

It is worth remembering that this measure is necessary, but temporary, and should be treated with patience and understanding. The child's initial discomfort goes away within about a week, then he gets used to it and no longer feels any discomfort from wearing the splint. The duration of use of such devices is determined by the doctor based on periodic examinations and ultrasound diagnostics.

Consequences and complications

Delayed diagnosis and lack/ineffective treatment can lead to a number of complications:

• Impaired functions of the hip joint.
• Shortening of the injured limb.
• Deformations of the glenoid cavity.
• Development of pelvic asymmetry and curvature of the spinal column.
• Formation of flexion-adduction contracture.
• Avascular post-reposition necrosis of the femoral head .
• Multiplanar deformations.
• Development of hip arthrosis / dysplastic coxarthrosis .

Ready solution to a genetics problem | Population genetics

Task 1. Congenital hip dislocation is inherited dominantly, the average penetrance of the gene is 25%. The disease occurs with a frequency of 6: 10,000 (V.P. Efroimson, 1968). Determine the number of homozygous individuals for the recessive gene. Solution: We formulate the problem condition in the form of a table:

Thus, from the conditions of the problem, according to the Hardy-Weinberg formula, we know the frequency of occurrence of genotypes AA and Aa, i.e. p2 + 2pq . It is necessary to find the frequency of occurrence of genotype aa, i.e. q 2. From the formula p2 + 2pq + q2 = 1 it is clear that the number of individuals homozygous for the recessive gene (aa) q2 = 1 - (p2 + 2pq). However, the number of patients given in the problem (6: 10,000) does not represent p2 + 2pq, but only 25% of gene A carriers, and the true number of people with this gene is four times larger, i.e. 24: 10,000. Therefore, p2 + 2pq = 24: 10,000 . Then q 2 (the number of individuals homozygous for the recessive gene) is equal to 1 - p2 + 2pq = 1 - 24: 10000 = 0.9976 or 9976: 10000.

Answer: The number of homozygous individuals for the recessive gene a is 9976:10000 or approximately 1:10.

Task 2 The Kidd blood group system is determined by the allelic genes Ik and Ik. The Ik gene is dominant to the Ik gene and individuals who have it are Kidd-positive. The frequency of the Ik gene among the population of Krakow is 0.458 (V. Socha, 1970). The frequency of Kidd-positive people among blacks is 80% (K. Stern, 1965). Determine the genetic structure of the population of Krakow and blacks according to the Kidd system. Solution: We formulate the problem condition in the form of a table:

We make a mathematical representation of the Hardy-Weinberg law, we get: p + q = 1, p2 + 2pq + q2 = 1. p is the frequency of occurrence of the Ik gene; q is the frequency of occurrence of the Ik gene; p2 — frequency of occurrence of dominant homozygotes (IkIk); 2pq—heterozygote frequency (Ik Ik); q2 is the frequency of occurrence of recessive homozygotes (IkIk). Thus, from the conditions of the problem, according to the Hardy-Weinberg formula, we know the frequency of occurrence of the dominant gene in the Krakow population - p = 0.458 (45.8%). We find the frequency of occurrence of the recessive gene: q = 1- 0.458 = 0.542 (54.2%). We calculate the genetic structure of the Krakow population: frequency of occurrence of dominant homozygotes - p2 = 0.2098 (20.98%); frequency of occurrence of heterozygotes - 2pq = 0.4965 (49.65%); frequency of occurrence of recessive homozygotes - q2 = 0.2937 (29.37%). For blacks, from the conditions of the problem, we know the frequency of occurrence of dominant homozygotes and heterozygotes (dominant trait), i.e. p2 + 2pq = 0.8. According to the Hardy-Weinberg formula, we find the frequency of occurrence of recessive homozygotes (IkIk): q2 = 1 - p2 + 2pq = 0.2 (20%). Now we calculate the frequency of the recessive gene Ik: q = 0.45 (45%). Find the frequency of occurrence of the Ik gene: p = 1-0.45 = 0.55 (55%); frequency of occurrence of dominant homozygotes ((IkIk): p2 = 0.3 (30%); frequency of occurrence of heterozygotes (IkIk): 2pq = 0.495 (49.5%).

Answer: 1. Genetic structure of the Krakow population according to the Kidd system: frequency of occurrence of dominant homozygotes (IkIk) - p2 = 0.2098 (20.98%); frequency of occurrence of heterozygotes - (IkIk) 2pq = 0.4965 (49.65%); frequency of occurrence of recessive homozygotes - (IkIk) q2 = 0.2937 (29.37%). 2. Genetic structure of the black population according to the Kidd system: frequency of occurrence of dominant homozygotes (IkIk) - p2 = 0.3 (30%); frequency of occurrence of heterozygotes - IkIk) 2pq = 0.495 (50%); frequency of occurrence of recessive homozygotes - (IkIk) q2 = 0.2 (20%).

Task 3 Tay-Sachs disease, caused by an autosomal recessive gene, is incurable; people suffering from this disease die in childhood. In one of the large populations, the frequency of births of sick children is 1: 5000. Will the concentration of the pathological gene and the frequency of this disease change in the next generation of this population? Solution. Solution: We formulate the problem condition in the form of a table:

We make a mathematical representation of the Hardy-Weinberg law p + q = 1, p2 + 2pq + q2 = 1. p is the frequency of occurrence of gene A; q is the frequency of occurrence of gene a; p2—frequency of occurrence of dominant homozygotes (AA); 2pq—frequency of occurrence of heterozygotes (Aa); q2 is the frequency of occurrence of recessive homozygotes (aa). From the conditions of the problem, according to the Hardy-Weinberg formula, we know the frequency of occurrence of sick children (aa), i.e. q2 = 1/5000. The gene causing this disease will pass to the next generation only from heterozygous parents, so it is necessary to find the frequency of occurrence of heterozygotes (Aa), i.e. 2pq. q = 1/71 = 0.014; p =1 - q = 1 - 0.014 = 0.986; 2pq = 2(0.986 * 0.014) = 0.028. We determine the gene concentration in the next generation. It will be in 50% of gametes in heterozygotes, its concentration in the gene pool is about 0.014. The probability of having sick children is q2 = 0.000196, or 0.000196/0.0002 = 0.98, i.e. 0.98 per 5000 population. Thus, the concentration of the pathological gene and the frequency of this disease in the next generation of this population will practically not change (there is a slight decrease).

Answer: The concentration of the pathological gene and the frequency of this disease in the next generation of this population will practically not change (according to the conditions of the problem - 1: 5000, and according to calculations - 0.98: 5000).

Task 4 The brown-eyed allele is dominant over blue-eyed. In a population, both alleles occur with equal probability. Father and mother are brown-eyed. What is the probability that the child born to them will be blue-eyed? Solution: Solution. If both alleles occur equally frequently in a population, then there are 1/4 (25%) dominant homozygotes, 1/2 (50%) heterozygotes (both brown-eyed) and 1/4 (25%) recessive homozygotes (blue-eyed) . Thus, if a person is brown-eyed, then it is two against one that he is a heterozygote, i.e. 75% heterozygotes and 25% homozygotes. So, the probability of being a heterozygote is 2/3. The probability of passing on the blue-eyed allele to offspring is 0 if the organism is homozygous, and 1/2 if it is heterozygous. The overall probability that a given brown-eyed parent will pass on the blue-eyed allele to their offspring is 2/3 . 1/2 = 1/3. For a child to be blue-eyed, he must receive an allele for blue eyes from each parent. This will happen with a probability of 1/3 . 1/3 = 1/9 (11.1%).

Answer: The probability of having a blue-eyed child from brown-eyed parents is 1/9.

Problem 5 Cystic fibrosis of the pancreas affects individuals with a recessive homozygous phenotype and occurs in the population with a frequency of 1 in 2000. Calculate the frequency of carriers of the cystic fibrosis gene. Solution: Carriers are heterozygotes. Genotype frequencies are calculated using the Hardy-Weinberg equation: p2 + 2pq + q2 = 1, where p2 is the frequency of the dominant homozygous genotype, 2pq is the frequency of the heterozygous genotype, q2 is the frequency of the recessive homozygous genotype. Cystic fibrosis of the pancreas affects individuals with a recessive homozygous phenotype; therefore, q2 = 1 in 2000, or 1/2000 = 0.0005. Hence q = = 0.0224 Since, p + q = 1; p = 1 – q = 1 – 0.0224 = 0.9776. Thus, the frequency of the heterozygous phenotype (2pq) = 2 . (0.9776) . (0.0224) = 0.044, i.e. carriers of the recessive gene for cystic fibrosis of the pancreas make up about 4.4% of the population.

Answer: The frequency of cystic fibrosis gene carriers is 4.4%.

Task 6 In the population there are three genotypes for the albinism gene a in the ratio: 9/16AA, 6/16Aa and 1/16aa. Is this population in a state of genetic equilibrium? Solution: Description of the karyotype: It is known that the population consists of 9/16AA, 6/16Aa and 1/16aa genotypes. Does this relationship correspond to the equilibrium in the population expressed by the Hardy-Weinberg formula? p2 + 2pq + q2 = 1. After transforming the numbers, it becomes clear that the population for a given characteristic is in a state of equilibrium: (3/4)2AA: 2 . 3/4 . 1/4Aa : (1/4)2 aa. Hence p = = 0.75; q = = 0.25. Which corresponds to the equation p + q = 1; 0.75 + 0.25 = 1. Answer: This population is in a state of genetic equilibrium.

Task 7 During a survey of one city with a population of 1,000,000 people, 49 albinos were discovered.BR. Determine the frequency of occurrence of heterozygous carriers of the albinism gene among residents of a given city. Solution: Since albinos are recessive homozygotes (aa), then, according to the Hardy-Weinberg law: p2 + 2pq + q2 = 1; q2 = 49/1000000 = 1/20408; the frequency of the recessive gene is: q2 = (1/20408)2. From which we get: q = 1/143; p + q = 1, hence p = 1 – q; p = 1 - 1/143 = 142/143. The frequency of heterozygotes is 2pq. 2pq = 2 . 142/143 . 1/143 = 284/20449 = 1/721/70.

Answer: Consequently, every 70th city resident is a heterozygous carrier of the albinism gene.

Problem 8 The population consists of 9% AA homozygotes, 42% Aa heterozygotes, 49% aa homozygotes. Determine the frequency of alleles A and a. Solution: Given: AA - 9%; Aa - 42%; aa - 49%. The total number of alleles in the population is 1 or 100%. AA homozygotes have only the A allele and their number is 9%, or 0.09 of the total number of alleles. Heterozygotes Aa make up 42: out of the total number of all individuals, or 0.42. They give 21%, or 0.21, of A alleles and the same amount (42%, or 0.21) of a alleles. The total number of A alleles will be 9% + 21% = 30%, or 0.3. Homozygotes aa carry 49%, or 0.49 a alleles. In addition, Aa heterozygotes produce 21%, or 0.21 a alleles, for a total of 49% + 21% = 70%, or 0.7. It follows that p = 0.09 + 0.21 = 0.3, or 30%; q = 0.49 + 0.21 = 0.7 or 70%.

Answer: p = 0.09 + 0.21 = 0.3, or 30%; q = 0.49 + 0.21 = 0.7 or 70%.

Task 9 Analysis of the population showed that the occurrence of people with an autosomal recessive trait is 0.04. What is the frequency of heterozygotes in this population? Solution: Given: 0.04 = q2; Need to find: 2pq. 1) q = = 0.2 2) p = 1 – q = 1 – 0.2 = 0.8 3) 2pq = 2 x 0.8 . 0.2 = 0.32.

Answer: The frequency of heterozygotes in this population is 0.32, or 32%.

Task 10 Rye albinism is a recessive trait. Among the 10,000 plants examined, 25 albino plants were found. Determine the % content of heterozygous plants. the discovered albino plants are aa homozygotes. Solution: Let's find the frequency of occurrence of these plants: q2 = 25/10000 = 0.0025. The frequency of occurrence of recessive alleles a will be: q = = 0.05. Since p + q = 1, then p = 1 - q = 1 - 0.05 = 0.95. Let's find the % content of heterozygous plants Aa: 2pq = 2(0.95 . 0.05) = 0.095, or 9.5%.

Answer: 9,5%.