Achondroplasia is the most common birth defect that is characterized by abnormal body proportions: people suffering from this disease have signs of relative macrocephaly (the forehead protrudes forward and the bridge of the nose is flat), the arms and legs are very short, while the legs and spine are deformed, the shoulders and hips the bones are shortened more than the radii, and the torso is of almost normal size.
Sometimes a large head is a reflection of hydrocephalus - the presence of excess fluid in the brain - and requires surgical intervention. The palms of patients with achondroplasia are small with short, thick fingers. There is some distance between the middle and ring fingers (trident hand). In most cases, people, having reached adulthood, have a height of no more than 120-130 cm.
What it is?
Achondroplasia is a genetic disease that results in impaired bone growth. This is one of the most common types of dwarfism. It is possible to judge whether a child has this pathology from the moment of birth.
Pathogenesis and genetics
Achondroplasia is caused by mutations in the FGFR3 gene, and is located on the fourth chromosome. At the same time, the growth of cartilage is inhibited. FGFR3 encodes a protein called Fibroblast Growth Factor 3. This protein is responsible for the growth of bones in the body. In achondroplasia, FGFR3 cannot function properly, and bone and cartilage growth slows down. This results in short bones, abnormal bone shape, and short stature.
FGFR3 is a transmembrane tyrosine kinase receptor that binds to FGF. The binding of FGF to the extracellular region of FGFR3 activates the intracellular domain of the receptor and triggers a signal sequence. In endochondral bone, FGFR3 activation inhibits chondrocyte proliferation in the growth plate, thereby helping to coordinate chondrocyte growth and differentiation with the growth and differentiation of bone progenitor cells.
Achondroplasia-associated FGFR3 mutations are gain-of-function mutations that cause ligand-independent activation of the FGFR3 protein. This constant upregulation of the FGFR3 protein improperly inhibits the proliferation of chondrocytes in the growth plate and leads to shortening of long bones, as well as abnormal formation of other bones.
Guanine at position 1138 in the FGFR3 gene is one of the most mutable nucleotides identified in all human genes. A mutation of this nucleotide occurs in almost 100% of cases of achondroplasia; more than 80% of patients have a new mutation. New guanine mutations at position 1138 of the FGFR3 gene occur exclusively in paternal germ cells and their frequency increases with paternal age (>35 years).
Symptoms of achondroplasia
Immediately after birth, you can detect the most obvious signs of achondroplasia in a newborn:
- Deep location of the eye orbits,
- The presence of additional folds on the inner corners of the eyes,
- Saddle-shaped flattened nose shape,
- Enlarged head
- Convex forehead
- Noticeably protruding parietal and occipital tubercles,
- Wide-set eyes
- Hydrocephalus,
- Strong protrusion of the buttocks due to the posterior position of the pelvis,
- Small chest
- Enlarged tonsils,
- Breathing problems
- Violation of muscle tone,
- Significant protrusion of the upper jaw,
- High sky
- Rough tongue
- Short arms and legs of the baby,
- Wide palms and feet with short toes,
- Retarded physical development
- Possible torticollis
- Hearing and vision impairments.
As they grow, all of these symptoms progress, leading to multiple complications, many of which can be fatal.
Diagnosis of achondroplasia
To assess the condition of organs and systems, a comprehensive examination is carried out. To exclude hydrocephalus in children with achondroplasia, MRI and CT scans of the brain are performed.
- When radiography of the skull in patients with achondroplasia, there is a disproportion between the brain and facial parts, the foramen magnum is reduced in size, the bones of the cranial vault and the lower jaw are enlarged. The Turkish saddle is characteristically shoe-shaped and flat, the base is elongated.
- X-rays of the spine usually do not reveal gross changes. Physiological curves are less pronounced than normal.
- X-ray of the chest is usually unchanged, in some cases the sternum protrudes forward and is slightly curved. Deformations and thickening of the ribs are possible. Sometimes the anatomical bends of the clavicle are absent.
- X-rays of tubular bones reveal shortening, thinning of the diaphyses, goblet expansion and thickening of the metaphyses.
- X-rays of the joints reveal deformation of the articular surfaces, disruption of the shape of the epiphyses, and widening of the joint spaces.
- X-ray of the pelvis reveals changes in the size and shape of the wings of the iliac bones. The horizontal location of the roof of the acetabulum is determined.
As a rule, making a diagnosis does not cause difficulties due to the characteristic appearance and proportions of the patient’s body.
Achondroplasia - treatment
Effective treatment of achondroplasia is not currently possible. Doctors can only minimize the consequences to improve the quality of life of patients.
In childhood, conservative therapy is carried out for such children - massage, physical therapy. This helps strengthen the muscle corset and prevent severe deformation of the lower extremities. Patients are advised to wear special shoes that reduce pressure on the bones. It is also important to follow a diet to avoid burdening the skeleton with excess weight. Defects in the jaw area are corrected by wearing special plates.
There are cases when hormone therapy is prescribed in childhood, which helps to slightly compensate for the lack of growth. Hormone therapy is not used for adults. If the parents decide to surgically solve the problem, then it is necessary to contact a specialized clinic that has sufficient experience in performing such operations.
- Surgery is performed if the disease causes obvious discomfort to the patient. An unambiguous decision in favor of surgery is made when there is a threat of spinal cord entrapment, kyphosis of the middle part of the back, or an “o”-shaped leg.
- It is also possible to lengthen the bones, for which several surgical interventions are performed in stages. At four to six years of age, children undergo surgery to lengthen their legs (up to six cm), hips (up to about seven to eight cm) and shoulders (possibly up to five centimeters). The duration of such stages is about five months with a break of two to three months. The next series of interventions is carried out at the age of fourteen to fifteen years. Here the patient also goes through three stages, the expected result is the same as the first time.
However, such actions do not completely eliminate the disease and its symptoms, since with short stature, elongation of bones even up to ten centimeters does not make patients look like ordinary people. In addition, not all patients are ready to undergo a series of operations and painful rehabilitation periods.
Inheritance of achondroplasia
For healthy parents of a child with achondroplasia, the risk of recurrence in future children is low, but probably somewhat higher than for the general population, since the possibility of sexual mosaicism, although extremely rare in achondroplasia, has been proven.
In a marriage where one partner has achondroplasia, the risk of recurrence in each child is 50%, since achondroplasia is an autosomal dominant disease with complete penetrance. In a marriage of two affected individuals, each child has a 50% risk of having achondroplasia, a 25% risk of lethal homozygous achondroplasia, and a 25% chance of normal growth.
When a mother with achondroplasia is pregnant, a fetus of normal size requires delivery by cesarean section.
Prevention
Prevention of achondroplasia consists of medical genetic consultation and prenatal diagnosis, which makes it possible to detect pathologies even at the stage of intrauterine development. Consultation with a geneticist is especially required for those who already have a history of dwarfism in their family. A special examination will allow you to assess the risk of having a sick child.
If parents already have achondroplasia, the disease cannot be prevented, as it is inherited.
Achondroplasia is a congenital disease associated with a genetic mutation, as a result of which a person’s legs are significantly reduced in length while maintaining normal body size. Signs of the disease include short stature - no more than 130 centimeters in adulthood, curvature of the spine, large head size with protruding frontal tubercles.
According to statistics, the disease can occur in one newborn out of 10 thousand; achondroplasia most often affects male newborns.
Achondroplasia occurs as a result of a mutation in the FGFR3 gene, which is directly responsible for the ossification and proliferation of cartilage tissue.
As a result of these problems, the limbs stop growing in length and acquire an abnormal structure. According to medical research, gene mutation occurs randomly and is practically independent of external factors.
The likelihood of a mutation increases in the following cases:
- The child’s parents (especially the father) are over 35 years of age;
- Achondroplasia was diagnosed in one of the baby’s parents.
The age of the parents is not considered a 100% factor in the development of achondroplasia in the unborn child.
Symptoms
A gene mutation can be judged without a diagnostic study: the baby has an overly large head and disproportionately small limbs, the frontal and occipital protuberances stand out on the head. Sometimes a newborn is diagnosed with hydrocephalus.
Other symptoms of the disease include:
- Disturbances in the normal structure of bones at the base of the head;
- There are folds in the eye area that are uncharacteristic for ordinary children; the eyes themselves are set too wide;
- The nose has a saddle-shaped appearance;
- Protruding upper jaw;
- Protruding frontal parts of the head;
- Raised palate and rough tongue.
The newborn has uniformly short limbs, they are curved in the articular areas. The child can only reach the umbilical area with his hands. The feet have an overly wide appearance and are shorter. The palms, like the feet, are wide, all fingers, except the first, are almost the same size.
A newborn with the mutation has numerous skin folds and fat deposits in the upper and lower extremities. Despite all the problems with growth, the patient’s torso develops correctly and is of normal size, changes do not affect the chest, while the stomach protrudes sharply forward, the pelvis tilts back, and the buttocks stand out strongly.
With achondroplasia, a child may experience problems with normal breathing, which are caused by features in the structure of the face and chest. Patients are often diagnosed with problems with normal physical development: the child begins to hold his head only after 3-4 months of life, the baby sits at 8-9 months from the day of birth, the child gets to his feet and begins to walk independently no earlier than 1.5- 2 years after birth. It is noteworthy that the lag in physical development is practically no different in intellectual abilities - the child’s mental development occurs within normal limits, he does not lag behind his peers.
As the child grows, the bones of the limbs continue to thicken and bend, and they acquire a lumpy structure. Subsequently, internal twisting of the femur bones in the lower sections begins, the knee joints acquire a loose appearance, and upon external examination, planovalgus feet are diagnosed.
The curvature affects not only the lower but also the upper limbs; the forearms are especially susceptible to distortion.
In adulthood, the height of women with a mutated gene does not exceed 124 centimeters, the height of adult men does not exceed 131 centimeters. All deformations of the head and facial skeleton that arise in childhood not only persist into adulthood, but also become more pronounced. Sometimes patients with achondroplasia develop strabismus, and due to problems with low mobility, patients develop obesity.
Achondroplasia is often accompanied by a decrease in the size of the spinal canal in the lumbar spine. This problem is accompanied by a decrease in sensitivity and the appearance of pain in the lower extremities. In the absence of proper treatment, narrowing of the spinal canal can result in paralysis of the legs and disruption of the normal functionality of internal organs in the pelvic area.
Diagnosis of the disease becomes possible without additional complex examinations due to the body structure and proportions of the patient characteristic of achondroplasia.
Typically, a sick child is sent for an additional comprehensive examination in order to determine the condition of the internal systems and organs:
Forms of the disease
Depending on the origin of the disease, it is customary to distinguish the following types of achondroplasia:
- Hereditary. The mutated gene is passed on from the affected parent to the child.
- Sporadic. The disease manifests itself regardless of hereditary factors; all mutations are spontaneous.
When a disease is detected, the patient is registered and observed by a doctor throughout his life from birth.
Treatment
Achondroplasia is a hereditary genetic disease for which there is no cure. In childhood, patients are prescribed conservative treatment, which is based on physical therapy and manual therapy. This approach allows not only to strengthen the muscles, but also to prevent deformation of the limbs.
Surgical intervention is indicated in advanced cases when patients have serious deformities in the skeletal structure. The following operations are carried out:
- Osteotomy – the leg bones are cut using a special tool and reattached in the correct position. The operation is used only in cases of severe deformation of the lower extremities; it allows you to slightly increase the length of the legs.
- Laminectomy - a dissection of the spinal canal is performed, thereby reducing the pressure. The operation is performed for pathologies in the structure of the spine.
If achondroplasia causes the development of concomitant diseases and complications, appropriate treatment is used to eliminate unpleasant symptoms.
Treatment also involves following a certain diet. Proper nutrition helps not only to avoid obesity, but also to ensure that the body receives essential vitamins and minerals.
Forecast
It is impossible to give a favorable prognosis when diagnosing achondroplasia in a child due to the fact that the disease cannot even be completely cured.
When carrying out therapeutic and preventive procedures immediately after the birth of a child, the prognosis is more favorable - all deformities can be reduced and the development of complications can be prevented.
The most unfavorable prognosis is given with gelatinous softening of the cartilage tissue - most newborns die either directly in the womb or a short time after birth.
Complications
The most common complication of achondroplasia is compression of the spinal canal and nerves.
Compression, in turn, can provoke:
- Problems in the normal functioning of the urinary system - patients often complain of urinary and fecal incontinence, problems with potency;
- The upper and lower limbs lose normal motor ability;
- Weakness occurs in the lower and upper extremities, and normal muscle tone is disrupted.
Prevention
Prevention of genetic disease is aimed at attending consultations in the field of medical genetics and undergoing prenatal diagnostic procedures for the fetus.
If one of the parents has a mutated gene, the child will most likely develop the disease, and it will be impossible to prevent it.
Preventive actions:
- Comprehensive examination when planning pregnancy;
- Diagnosis of the disease before the birth of the child;
- Mandatory prenatal examinations.
Achondroplasia is a serious genetic disease from which it is completely impossible to cure. Remember that if you start treatment in childhood, most complications can be avoided.
Within the gene pool of a population, the proportion of genotypes containing different alleles of the same gene; subject to certain conditions, it does not change from generation to generation. These conditions are described by the basic law of population genetics, formulated in 1908 by the English mathematician J. Hardy and the German geneticist G. Weinberg. “In a population of an infinitely large number of freely interbreeding individuals, in the absence of mutation, selective migration of organisms with different genotypes, and the pressure of natural selection, the original allele frequencies are maintained from generation to generation.”
Hardy-Weinberg equation in solving genetic problems
It is well known that this law is applicable only for ideal populations: a sufficiently high number of individuals in the population; the population must be panmixed, when there are no restrictions on the free choice of a sexual partner; there should be practically no mutation of the trait being studied; there is no influx and outflow of genes and there is no natural selection.
The Hardy-Weinberg law is formulated as follows:
in an ideal population, the ratio of the frequencies of gene alleles and genotypes from generation to generation is a constant value and corresponds to the equation:
p 2 +2pq + q 2 = 1
Where p 2 is the proportion of homozygotes for one of the alleles; p is the frequency of this allele; q 2 is the proportion of homozygotes for the alternative allele; q is the frequency of the corresponding allele; 2pq—proportion of heterozygotes.
What does it mean “the ratio of gene allele frequencies” and “the ratio of genotypes” - constant values? What are these values?
Let the frequency of occurrence of a gene in a dominant state (A) be equal to p, and the frequency of a recessive allele (a) of the same gene is equal to q(it’s possible vice versa, or even using one letter, expressing one designation from another) and understanding that the sum of the frequencies of the dominant and recessive alleles of one gene in a population is equal to 1, we get the first equation:
1) p + q = 1
Where does the Hardy-Weinberg equation itself come from? You remember that when monohybrid crossing heterozygous organisms with genotypes Aa x Aa according to Mendel’s second law, we will observe the appearance of different genotypes in the offspring in the ratio 1AA: 2 Aa: 1aa.
Since the frequency of occurrence of the dominant allelic gene A is designated by the letter p, and the recessive allele a by the letter q, the sum of the frequencies of occurrence of the genotypes of organisms themselves (AA, 2Aa and aa) having the same allelic genes A and a will also be equal to 1, then :
2) p 2 AA + 2pqAa + q 2 aa =1
In population genetics problems, as a rule, the following is required:
a) find the frequency of occurrence of each of the allelic genes based on the known ratio of the frequencies of genotypes of individuals;
B) or vice versa, find the frequency of occurrence of any of the genotypes of individuals based on the known frequency of occurrence of the dominant or recessive allele of the trait being studied.
So, by substituting the known value of the frequency of occurrence of one of the alleles of a gene into the first formula and finding the value of the frequency of occurrence of the second allele, we can always use the Hardy-Weinberg equation to find the frequency of occurrence of the various genotypes of the offspring themselves.
Usually some actions (due to their obviousness) are decided in the mind. But in order to make clear what is already obvious, you need to understand well what the letter designations in the Hardy-Weinberg formula are.
The provisions of the Hardy-Weinberg law also apply to multiple alleles. Thus, if an autosomal gene is represented by three alleles (A, a1 and a2), then the formulas of the law take on the following form:
RA + qa1 + ra2 = 1;
P 2 AA+ q 2 a1a1 + r 2 a2a2 + 2pqAa1 + 2prAa2 + 2qra1a2 = 1.
"In a population from an infinite number of freely interbreeding individuals V absence of mutations, selective migration organisms with different genotypes and pressure of natural selection original allele frequencies are maintained from generation to generation.”
Let us assume that in the gene pool of a population that satisfies the described conditions, a certain gene is represented by alleles A 1 and A 2, found with a frequency of p and q. Since there are no other alleles in this gene pool, then p + q = 1. In this case, q = 1 - p.
Accordingly, individuals of a given population form p gametes with the A 1 allele and q gametes with the A 2 allele. If crossings occur randomly, then the proportion of germ cells connecting with gametes A 1 is equal to p, and the proportion of germ cells connecting with gametes A 2 is q. Generation F 1, which arises as a result of the described reproduction cycle, is formed by the genotypes A l A 1, A 1 A 2, A 2 A 2, the number of which is correlated as (p + q) (p + q) = p 2 + 2pq + q 2 (Fig. 10.2). Upon reaching sexual maturity, individuals AlAi and ArA2 each form one type of gamete - A 1 or A 2 - with a frequency proportional to the number of organisms of the indicated genotypes (p and q). Individuals A 1 A 2 form both types of gametes with an equal frequency 2pq /2.
Rice. Regular distribution of genotypes in a series of generations depending on the frequency of formation of gametes of different types (Hardy-Weinberg law)
Thus, the proportion of gametes A 1 in generation F 1 will be p 2 + 2pq/2 = p 2 + p(1-p) = p, and the proportion of gametes A 2 will be equal to q 2 + 2pq/2 = q 2 + + q (l -q) = q.
Since the frequencies of gametes with different alleles in generation fi are not changed in comparison with the parent generation, generation F 2 will be represented by organisms with genotypes A l A 1, A 1 A 2 and A 2 A 2 in the same ratio p 2 + 2pq + q 2 . Thanks to this, the next cycle of reproduction will occur in the presence of p gametes A 1 and q gametes A 2. Similar calculations can be made for loci with any number of alleles. The conservation of allele frequencies is based on statistical patterns of random events in large samples.
The Hardy-Weinberg equation, as discussed above, is valid for autosomal genes. For sex-linked genes, the equilibrium frequencies of genotypes A l A 1, A 1 A 2 and A 2 A 2 coincide with those for autosomal genes: p 2 + 2pq + q 2. For males (in the case of heterogametic sex), due to their hemizygosity, only two genotypes A 1 - or A 2 - are possible, which are reproduced with a frequency equal to the frequency of the corresponding alleles in females in the previous generation: p and q. It follows that phenotypes determined by recessive alleles of genes linked to chromosome X are more common in males than in females.
Thus, with a hemophilia allele frequency of 0.0001, this disease is observed in men of this population 10,000 times more often than in women (1 in 10 thousand in the former and 1 in 100 million in the latter).
Another general consequence is that in the case of inequality of allele frequencies in males and females, the difference between the frequencies in the next generation is halved, and the sign of this difference changes. It usually takes several generations for the frequencies to reach equilibrium in both sexes. The specified state for autosomal genes is achieved in one generation.
The Hardy-Weinberg law describes the conditions genetic stability of the population. A population whose gene pool does not change over generations is called Mendelian. The genetic stability of Mendelian populations puts them outside the process of evolution, since under such conditions the action of natural selection is suspended. The identification of Mendelian populations is of purely theoretical significance. These populations do not occur in nature. The Hardy-Weinberg law lists conditions that naturally change the gene pools of populations. This result is led, for example, by factors limiting free crossing (panmixia), such as the finite number of organisms in the population, isolation barriers that prevent the random selection of mating pairs. Genetic inertia is also overcome through mutations, the influx into or outflow of individuals with certain genotypes into a population, and selection.
Examples of solutions to some tasks using the Hardy-Weinberg equation.
Problem 1. In the human population, the number of individuals with brown eyes is 51%, and with blue eyes - 49%. Determine the percentage of dominant homozygotes in this population.
The difficulty of solving such tasks lies in their apparent simplicity. Since there is so little data, then the solution should seem to be very short. It turns out not very much.
According to the conditions of this kind of task, we are usually given information about the total number of phenotypes of individuals in the population. Since the phenotypes of individuals in a population with dominant traits can be represented by both individuals homozygous for the genotype AA and heterozygous Aa, then in order to determine the frequencies of occurrence of any specific genotypes of individuals in this population, it is necessary to first calculate the frequencies of alleles of the A and a genes separately .
How should we reason when solving this problem?
Since it is known that brown eye color is dominant over blue, we will designate the allele responsible for the manifestation of the brown-eyed trait as A, and the allele gene responsible for the manifestation of blue eyes, respectively, as a. Then the brown-eyed people in the population under study will be people with both the AA genotype (dominant homozygotes, the proportion of which must be found according to the conditions of the problem) and Aa heterozygotes), and the blue-eyed people will be only aa (recessive homozygotes).
According to the conditions of the problem, we know that the number of people with the AA and Aa genotypes is 51%, and the number of people with the aa genotype is 49%. How, based on these statistics (there should be a large, representative sample), can one calculate the percentage of brown-eyed people with only the AA genotype?
To do this, let us calculate the frequency of occurrence of each of the allelic genes A and a in a given population of people. The Hardy-Weinberg law, applied to large, freely interbreeding populations, will allow us to do just that.
Having designated the frequency of occurrence of allele A in a given population by the letter q, we have the frequency of occurrence of the allele gene a = 1 - q. (It would be possible to indicate the frequency of occurrence of the allelic gene a with a separate letter, as in the text above - this is more convenient for everyone). Then the Hardy-Weinberg formula for calculating genotype frequencies in monohybrid crossings with complete dominance of one allelic gene over the other will look like this:
q 2 AA+ 2q(1 - q)Aa + (1 - q) 2 aa = 1.
Well, now everything is simple, you probably all guessed what we know in this equation, and what should be found?
(1 - q) 2 = 0.49 is the frequency of occurrence of people with blue eyes.
Find the value of q: 1 - q = square root of 0.49 = 0.7; q = 1 - 0.7 = 0.3, then q2 = 0.09.
This means that the frequency of brown-eyed homozygous AA individuals in this population will be 0.09 or their proportion will be 9%.
Task 2. In red clover, late ripeness dominates over early ripeness and is inherited monogenically. During testing, it was found that 4% of the plants belong to the early-ripening type of clover; what proportion of the late-ripening plants are heterozygotes?
In this context, approbation means assessing the purity of a variety. But isn’t a variety a pure line, like Mendel’s pea varieties, for example? Theoretically, “yes,” but in practice (the fields are large - these are not the experimental plots of the brilliant Mendel) in each production variety there may be some amount of “junk” gene alleles.
In this case, with a late-ripening clover variety, if the variety were pure, only plants with the AA genotype would be present. But the variety turned out to be not very pure at the time of testing (approbation), since 4% of the individuals were early ripening plants with the aa genotype. This means that alleles “a” have been included in this variety.
So, since they are “wormed in”, then in this variety there should also be individuals, although late-ripening in phenotype, but heterozygous with the Aa genotype - do we need to determine their number?
According to the conditions of the problem, 4% of individuals with the aa genotype will constitute 0.04 of the entire variety. In fact, this is q 2, which means the frequency of occurrence of the recessive allele a is q = 0.2. Then the frequency of occurrence of the dominant allele A is p = 1 - 0.2 = 0.8.
Hence the number of late-ripening homozygotes p2 = 0.64 or 64%. Then the number of Aa heterozygotes will be 100% - 4% - 64% = 32%. Since the total number of late-ripening plants is 96%, the proportion of heterozygotes among them will be: 32 x 100: 96 = 33.3%.
Problem 3. Using the Hardy-Weinberg formula for incomplete dominance
When examining the population of Karakul sheep, 729 long-eared individuals (AA), 111 short-eared individuals (Aa) and 4 earless individuals (aa) were identified. Calculate the observed phenotype frequencies, allele frequencies, and expected genotype frequencies using the Hardy-Weinberg formula.
This is an incomplete dominance problem, therefore, the frequency distributions of genotypes and phenotypes coincide and could be determined based on the available data. To do this, you simply need to find the sum of all individuals of the population (it is equal to 844), find the proportion of long-eared, short-eared and earless, first in percentage (86.37, 13.15 and 0.47, respectively) and in frequency shares (0.8637, 0.1315 and 0.00474).
But the task says to apply the Hardy-Weinberg formula to calculate genotypes and phenotypes and, in addition, to calculate the frequencies of alleles of genes A and a. So, to calculate the gene allele frequencies themselves, you cannot do without the Hardy-Weinberg formula.
Please note that in this task, unlike the previous one, to designate the frequencies of allelic genes, we will use the notation not as in the first task, but as discussed above in the text. It is clear that the result will not change, but you will have the right in the future to use any of these notation methods, whichever seems more convenient to you for understanding and carrying out the calculations themselves.
Let us denote the frequency of occurrence of allele A in all gametes of a sheep population by the letter p, and the frequency of occurrence of allele a by the letter q. Remember that the sum of allelic gene frequencies p + q = 1.
Since, according to the Hardy-Weinberg formula p 2 AA + 2pqAa + q 2 aa = 1, we have that the frequency of occurrence of earless q2 is equal to 0.00474, then by taking the square root of the number 0.00474 we find the frequency of occurrence of the recessive allele a. It is equal to 0.06884.
From here we can find the frequency of occurrence of the dominant allele A. It is equal to 1 - 0.06884 = 0.93116.
Now, using the formula, we can again calculate the frequencies of occurrence of long-eared (AA), earless (aa) and short-eared (Aa) individuals. Long-eared with the AA genotype will have p 2 = 0.931162 = 0.86706, earless with the aa genotype will have q 2 = 0.00474 and short-eared with the Aa genotype will have 2pq = 0.12820. (The newly obtained numbers calculated using the formula almost coincide with those calculated initially, which indicates the validity of the Hardy-Weinberg law).
Problem 4. Why is the proportion of albinos in populations so small
In a sample of 84,000 rye plants, 210 plants turned out to be albino, because... their recessive genes are in a homozygous state. Determine the frequencies of alleles A and a, as well as the frequency of heterozygous plants.
Let us denote the frequency of occurrence of the dominant allelic gene A by the letter p, and the frequency of the recessive gene a by the letter q. Then what can the Hardy-Weinberg formula p 2 AA + 2pqAa + q 2 aa = 1 give us for applying it to this problem?
Since we know the total number of all individuals of this rye population is 84,000 plants, and in parts this is 1, then the proportion of homozygous albino individuals with genotype aa equal to q2, of which there are only 210 pieces, will be q2 = 210: 84000 = 0.0025, then q = 0.05; p = 1 - q = 0.95 and then 2pq = 0.095.
Answer: allele frequency a - 0.05; allele A frequency - 0.95; the frequency of heterozygous plants with genotype Aa will be 0.095.
Problem 5. We raised chinchilla rabbits and ended up with albino rabbits.
In rabbits, the chinchilla hair color (Cch gene) dominates over albinism (Ca gene). CchCa heterozygotes are light gray in color. Albinos appeared among the young chinchilla rabbits on a rabbit farm. Of the 5,400 rabbits, 17 turned out to be albinos. Using the Hardy-Weinberg formula, determine how many homozygous rabbits with chinchilla coloring were obtained.
Do you think that the resulting sample of 5400 rabbits in the rabbit population can allow us to use the Hardy-Weinberg formula? Yes, the sample is significant, the population is isolated (rabbit farm) and the Hardy-Weinberg formula can indeed be applied in the calculations. To use it correctly, we must clearly understand what is given to us and what needs to be found.
Just for convenience of design, we will designate the genotype of chinchillas as AA (we will need to determine the number of them), the genotype of albino aa, then the genotype of heterozygous grays will be designated Aa.
If you “add up” all the rabbits with different genotypes in the studied population: AA + Aa + aa, then this will be a total of 5400 individuals.
Moreover, we know that there were 17 rabbits with the aa genotype. How can we now, without knowing how many heterozygous gray rabbits with the Aa genotype there were, determine how many chinchillas with the AA genotype are in this population?
As we can see, this task is almost a “copy” of the first one, only there we were given the results of calculations in the human population of brown-eyed and blue-eyed individuals in %, but here we actually know the number of albino rabbits - 17 pieces and all homozygous chinchillas and heterozygous graylings in total : 5400 - 17 = 5383 pieces.
Let's take 5400 pieces of all rabbits as 100%, then 5383 rabbits (the sum of genotypes AA and Aa) will be 99.685% or in parts it will be 0.99685.
Q 2 + 2q(1 - q) = 0.99685 is the frequency of occurrence of all chinchillas, both homozygous (AA) and heterozygous (Aa).
Then from the Hardy-Weinberg equation: q2 AA+ 2q(1 - q)Aa + (1 - q)2aa = 1, we find
(1 - q) 2 = 1 - 0.99685 = 0.00315 is the frequency of occurrence of albino rabbits with the aa genotype. Find what the value 1 - q is equal to. This is the square root of 0.00315 = 0.056. And q then equals 0.944.
Q 2 is equal to 0.891, and this is the proportion of homozygous chinchillas with the AA genotype. Since this percentage value will be 89.1% of 5400 individuals, the number of homozygous chinchillas will be 4811 pieces.
Task 6. Determining the frequency of occurrence of heterozygous individuals based on the known frequency of occurrence of recessive homozygotes
One form of glycosuria is inherited as an autosomal recessive trait and occurs with a frequency of 7:1000000. Determine the frequency of occurrence of heterozygotes in the population.
Let us designate the allelic gene responsible for the manifestation of glycosuria a, since it is said that this disease is inherited as a recessive trait. Then the allelic dominant gene responsible for the absence of the disease will be denoted by A.
Healthy individuals in the human population have genotypes AA and Aa; sick individuals have only the aa genotype.
Let us denote the frequency of occurrence of the recessive allele a by the letter q, and the frequency of the dominant allele A by the letter p.
Since we know that the frequency of occurrence of sick people with the aa genotype (which means q 2) is 0.000007, then q = 0.00264575
Since p + q = 1, then p = 1 - q = 0.9973543, and p2 = 0.9947155
Now substituting the values of p and q into the formula:
P2AA + 2pqAa + q2aa = 1,
Let's find the frequency of occurrence of heterozygous 2pq individuals in the human population:
2pq = 1 - p 2 - q 2 = 1 - 0.9947155 - 0.000007 = 0.0052775.
Task 7. Like the previous task, but about albinism
General albinism (milky white skin color, lack of melanin in the skin, hair follicles and retinal epithelium) is inherited as a recessive autosomal trait. The disease occurs with a frequency of 1: 20,000 (K. Stern, 1965). Determine the percentage of heterozygous gene carriers.
Since this trait is recessive, sick organisms will have the aa genotype - their frequency is 1: 20,000 or 0.00005.
The frequency of allele a will be the square root of this number, that is, 0.0071. The frequency of the A allele will be 1 - 0.0071 = 0.9929, and the frequency of healthy AA homozygotes will be 0.9859.
The frequency of all heterozygotes 2Aa = 1 - (AA + aa) = 0.014 or 1.4%.
Problem 8. Everything seems so simple when you know how to solve it
According to the Rh blood group system, the European population contains 85% Rh positive individuals. Determine the saturation of the population with the recessive allele.
We know that the allelic gene responsible for the manifestation of Rh positive blood is dominant R (let's denote its frequency of occurrence by the letter p), and Rh negative is recessive r (let's denote its frequency of occurrence by the letter q).
Since the problem says that p 2 RR + 2pqRr accounts for 85% of people, this means that Rh-negative phenotypes q 2 rr will account for 15% or their frequency of occurrence will be 0.15 of all people in the European population.
Then the frequency of occurrence of the r allele or “the saturation of the population with the recessive allele” (denoted by the letter q) will be the square root of 0.15 = 0.39 or 39%.
Task 9. The main thing is to know what penetrance is
Congenital hip dislocation is inherited dominantly. The average penetrance is 25%. The disease occurs with a frequency of 6:10,000. Determine the number of homozygous individuals in the population for a recessive trait.
Penetrance is a quantitative indicator of the phenotypic variability of gene expression.
Penetrance is measured as the percentage of the number of individuals in which a given gene manifested itself in the phenotype to the total number of individuals in whose genotype this gene is present in the state necessary for its manifestation (homozygous in the case of recessive genes or heterozygous in the case of dominant genes). The manifestation of a gene in 100% of individuals with the corresponding genotype is called complete penetrance, and in other cases - incomplete penetrance.
The dominant allele is responsible for the trait being studied, let’s denote it A. This means that organisms with this disease have genotypes AA and Aa.
It is known that phenotypically hip dislocation is detected in 6 organisms out of the entire population (10,000 examined), but this is only one fourth of all people who actually have genotypes AA and Aa (since it is said that the penetrance is 25%).
This means that in fact there are 4 times more people with genotypes AA and Aa, that is, 24 out of 10,000 or 0.0024. Then there will be 1 - 0.0024 = 0.9976 people with genotype aa, or 9976 people out of 10,000.
Problem 10. If only men get sick
Gout occurs in 2% of people and is caused by an autosomal dominant gene. In women, the gout gene does not manifest itself; in men, its penetrance is 20% (V.P. Efroimson, 1968). Determine the genetic structure of the population based on the analyzed trait based on these data.
Since gout is detected in 2% of men, that is, in 2 people out of 100 with a penetrance of 20%, then 5 times more men, that is, 10 people out of 100, are actually carriers of the gout genes.
But, since men make up only half of the population, then in total there will be 5 people out of 100 people with genotypes AA + 2Aa in the population, which means 95 out of 100 will have the aa genotype.
If the frequency of occurrence of organisms with genotypes aa is 0.95, then the frequency of occurrence of the recessive allele a in this population is equal to the square root of 0.95 = 0.975. Then the frequency of occurrence of the dominant allele “A” in this population is 1 - 0.975 = 0.005.
Task 11. How few people are resistant to HIV infection
Resistance to HIV infection is associated with the presence of certain recessive genes in the genotype, for example, CCR and SRF. The frequency of the recessive allele CCR-5 in the Russian population is 0.25%, and the allele SRF is 0.05%. In the Kazakh population, the frequency of these alleles is 0.12% and 0.1%, respectively. Calculate the frequencies of organisms that have increased resistance to HIV infection in each population.
It is clear that only homozygous organisms with the aa genotypes will have increased resistance to HIV infection. Organisms with genotypes AA (homozygotes) or Aa (heterozygotes) are not resistant to HIV infection.
In the Russian population of resistant organisms, the CCR allelic gene will be O.25% squared = 0.0625%, and the SRF allelic gene will be 0.05% squared = 0.0025%.
In the Kazakh population of resistant organisms, the CCR allelic gene will be O.12% squared = 0.0144%, and the SRF allelic gene will be 0.1% squared = 0.01%.
Population genetics
Solving typical problems
Problem 1 . The South American jungle is home to an aboriginal population of 127 people (including children). The frequency of blood group M is 64%. Is it possible to calculate the frequencies of blood group N and MN in this population?
Solution . For a small population, the mathematical expression of the Hardy-Weinberg law cannot be applied, so it is impossible to calculate gene frequencies.
Task 2. Tay-Sachs disease, caused by an autosomal recessive gene, is incurable; people suffering from this disease die in childhood. In one large population, the birth rate of affected 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
|
Sign |
Gene |
Genotype |
|
Tay-Sachs disease |
ahh |
|
|
Norm |
We make a mathematical notation of the Hardy-Weinberg law
p + q - 1, p 2 .+ 2 pq + q 2 = 1.
p frequency of occurrence of gene A;
q frequency of occurrence of gene a;
p 2 frequency of occurrence of dominant homozygotes
(AA);
2 pq frequency of occurrence of heterozygotes (Aa);
q 2 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. q 2 = 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. 2 pq.
q = 1/71, p = l - q - 70/71, 2 pq = 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. Probability of having sick children q 2 = 0.000196, or 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 (the decrease is insignificant).
Task 3. Congenital hip dislocation is inherited dominantly, the average penetrance of the gene is 25%. The disease occurs with a frequency of 6:10000 (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:
|
Sign |
Gene |
Genotype |
|
Norm |
ahh |
|
|
Hip dislocation |
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. p 2 + 2 pq . It is necessary to find the frequency of occurrence of genotype aa, i.e. q2.
From the formula p 2 - t - 2 pq + q 2 = l it is clear that the number of individuals homozygous for the recessive gene (aa) q 2 = 1 (p 2 + 2 pq ). However, the number of patients given in the problem (6:10,000) is not p 2 + 2 pq , but only 25% are carriers of gene A, but the true number of people who have this gene is four times larger, i.e. 24: 10,000. Therefore, p 2 + 2 pq = 24:10 000. Then q 2 (the number
individuals homozygous for the recessive gene) is 9976:10,000.
Problem 4 . The Kidd blood group system is determined by allelic genes Ik a and Ik in . Gene Ik a is dominant to the gene Ik in and individuals who have it are Kidd positive. Gene frequency Ik a among the population of Krakow is 0.458 (W. 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:
|
Sign |
Gene |
Genotype |
|
Kidz positive blood |
Ik α |
lk α lk α ;lk β lk β . |
|
Kidd negative blood |
Ik β |
Ik β Ik β |
We make a mathematical notation of the Hardy-Weinberg law: - p + q = I, p 2 + 2 pq + q 2 = 1.
p gene frequency Ik α ;
q gene frequency Ik β ; . p 2 frequency of occurrence of dominant homozygotes ( Ik α lk α );
2 pq frequency of occurrence of heterozygotes ( Ik α Ik β );
q 2 frequency of occurrence of recessive homozygotes ( Ik β Ik β ).
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%). 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 p 2 = 0.2098 (20.98%); frequency of occurrence of heterozygotes 2 pq = 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 (with
dominant sign), i.e. R 2 +2 pq =0.8. According to the Hardy-Weinberg formula, we find the frequency of occurrence of recessive homozygotes ( Ik β Ik β ): q 2 =1р 2 +2 pq =0.2 (20%). Now we determine the frequency of the recessive gene Ik β : q =0.45 (45%). Finding the frequency of occurrence of a gene Ik α : p=1-0.45=0.55 (55%); frequency of occurrence of dominant homozygotes ( Ik α Ik α ): р 2 = 0.3 (30%); frequency of occurrence of heterozygotes ( Ik α Ik β ): 2 pq = 0.495 (49.5%).
SELF-CONTROL TASKS
Problem 1 . Children with phenylketonuria are born with a frequency of 1:10,000 newborns. Determine the percentage of heterozygous gene carriers.
Problem 2 . General albinism (milky white skin color, lack of melanin in the skin, hair follicles and retinal epithelium) is inherited as a recessive autosomal trait. The disease occurs with a frequency of 1: 20,000 (K. Stern, 1965). Determine the percentage of heterozygous gene carriers.
Problem 3 . Hereditary methemoglobinemia, an autosomal recessive trait, occurs in Alaskan Eskimos with a frequency of 0.09%. Determine the genetic structure of the population for this trait.
Problem 4 . People with blood type N make up 16% of the Ukrainian population. Determine the frequency of groups M and MN.
Problem 5 . Papuans have a frequency of N blood group of 81%. Determine the frequency of groups M and MN in this population.
Task 6. During a survey of the population of southern Poland, individuals with blood groups were found: M 11163, MN 15267, N 5134. Determine the frequency of genes L N and L M among the population of southern Poland.
Problem 7 . The incidence of gout is 2%; it is caused by a dominant autosomal gene. According to some data (V.P. Efroimson, 1968), the penetrance of the gout gene in men is 20%, and in women 0%.
Determine the genetic structure of the population based on the analyzed trait.
Task 8. In the United States, about 30% of the population perceives the bitter taste of phenylthiourea (PTC), while 70% do not. The ability to taste FTC is determined by the recessive gene a. Determine the frequency of alleles A and a in this population.
Problem 9 . One of the forms of fructosuria is inherited as an autosomal recessive trait and occurs with a frequency of 7: 1,000,000 (V.P. Efroimson, 1968). Determine the frequency of heterozygotes in the population.
Problem 10. Determine the frequency of occurrence of albinos in a large African population, where the concentration of the pathological recessive gene is 10%.
Problem 11. Aniridia (absence of the iris) is inherited as an autosomal dominant trait and occurs with a frequency of 1:10,000 (V.P. Efroimson, 1968). Determine the frequency of heterozygotes in the population.
Problem 12 . Essential pentosuria (excretion in urine L -xylulose) is inherited as an autosomal recessive trait and occurs with a frequency of 1: 50,000 (L.O. Badalyan, 1971). Determine the frequency of occurrence of dominant homozygotes in the population.
Problem 13. Alkaptonuria (excretion of homogentisic acid in the urine, staining of cartilage tissue, development of arthritis) is inherited as an autosomal recessive trait with a frequency of 1:100,000 (V.P. Efroimson, 1968). Determine the frequency of heterozygotes in the population.
Problem 14 . Blood groups according to the M and N antigen system (M, MN, N ) are determined by codominant genes L N and L M . Gene frequency L M among the white population of the USA it is 54%, among the Indians 78%, among the Eskimos of Greenland 91%, among the Australian aborigines 18%. Determine the frequency of occurrence of blood groups MN in each of these populations.
Problem 15 . One grain of wheat, heterozygous for gene A, accidentally fell on a desert island. The grain sprouted and gave rise to a series of generations reproducing by self-pollination. What will be the proportion of heterozygous plants among representatives of the first, second, third; fourth generations, if the trait determined by the gene does not affect the survival of plants and their reproduction?
Problem 16 . Albinism in rye is inherited as an autosomal recessive trait. In the surveyed area, 210 albinos were found among 84,000 plants. Determine the frequency of occurrence of the albinism gene in rye.
Problem 17 . On one of the islands, 10,000 foxes were shot. 9991 of them turned out to be red (dominant trait) and 9 individuals were white (recessive trait). Determine the frequency of occurrence of the genotypes of homozygous red foxes, heterozygous red and white foxes in this population.
Problem 18. In a large population, the frequency of the color blindness gene (recessive, linked to X -chromosomal trait) among men is 0.08. Determine the frequency of occurrence of genotypes of dominant homozygotes, heterozygotes, and recessive homozygotes in women of this population.
Problem 19 . In Short Horn cattle, color is inherited as an autosomal trait with incomplete dominance: hybrids from crossing red and white animals have a roan color. Near N , which specializes in shorthorn breeding, has registered 4,169 reds, 3,780 roans and 756 whites. Determine the frequency of genes that determine the red and white coloration of livestock in a given area.
- HUMAN GENETICS
SOLVING TYPICAL PROBLEMS
Task 1. Define inheritance type
Solution. The trait occurs in every generation. This immediately excludes the recessive type of inheritance. Since this trait occurs in both men and women, this excludes the holandric type of inheritance. This leaves two possible modes of inheritance: autosomal dominant and sex-linked dominant, which are very similar. In a man II 3 have daughters with this trait ( III 1, III 5, III 7), and without it ( III -3), which excludes a sex-linked dominant type of inheritance. This means that this pedigree has an autosomal dominant type of inheritance.
Problem 2 
Solution. The trait does not occur in every generation. This excludes the dominant type of inheritance. Since the trait occurs in both men and women, this excludes the holandric type of inheritance. To exclude a sex-linked recessive type of inheritance, it is necessary to consider the marriage scheme Ш3 and III 4 (the sign does not occur in men and women). If we assume that the genotype of a man X A Y , and the genotype of the woman is X A X a , they cannot have a daughter with this trait (X a X a ), and in this pedigree there is a daughter with this trait IV -2. Considering the occurrence of the trait equally in men and women and the case of consanguineous marriage, we can conclude that an autosomal recessive type of inheritance occurs in this pedigree.
Task 3. The concordance of monozygotic twins by body weight is 80%, and that of dizygotic twins is 30%. What is the relationship between hereditary and environmental factors in the formation of a trait?
Solution. Using the Holzinger formula, we calculate the heritability coefficient:
KMB%-KDB%
100%-KDB%
N =
80% - 30%
100%-30%
Since the heritability coefficient is 0.71, the genotype plays a large role in the formation of the trait.
SELF-CONTROL TASKS
Problem 1 . Determine the type of inheritance.
Problem 2 . Determine the type of inheritance.
Problem 3 . Determine the type of inheritance.
Task 4. Blood groups according to the ABO system in monozygotes
In 100% of cases, identical twins coincide, and in 40% of dizygotic twins. What determines the coefficient of heritability - environment or heredity?
Problem 5 . Vitamin resistant rickets D (hypophosphatemia) is a hereditary disease caused by a dominant gene localized on the X chromosome. In a family where the father suffers from this disease and the mother is healthy, there are 3 daughters and 3 sons. How many of them could be sick?
Task 6. Is the composition of proteins the same in two monozygotic twins if there are no mutations in their cells?
Task 7. Which of the following characteristics characterize the autosomal dominant type of inheritance: a) the disease is equally common in women and men; b) the disease is transmitted from parents to children in each generation; c) a sick father has all his daughters sick; d) a son never inherits a disease from his father; d) are the parents of the sick child healthy?
Task 8. Which of the following characteristics characterize the autosomal recessive type of inheritance: a) the disease is equally common in women and men; b) the disease is transmitted from parents to children in each generation; c) a sick father has all his daughters sick; d) parents are blood relatives; d) are the parents of the sick child healthy?
Problem 9 . Which of the following characteristics characterize the dominant, X-linked type of inheritance: a) the disease is equally common in women and men; b) the disease is transmitted from parents to children in each generation; c) a sick father has all his daughters sick; d) a son never inherits a disease from his father; e) if the mother is sick, then, regardless of gender, the probability of having a sick child is 50%?
No one can be protected from the birth of a sick child. This is a big blow not only for parents, but also for medical staff. This is especially true for pathologies that cannot be prevented or prevented. Achondroplasia is one of these diseases.
Achondroplasia is a genetic disease, the main characteristic of which is short limbs, while the body length remains normal. The average height of the patient is 130 centimeters, and in some cases even less. The spine of such a patient has a curved shape, the head is large, and the frontal tubercles protrude significantly.
The incidence of pathology among all newborns is 1:10,000, and this affects girls more often than boys.
This pathology cannot be cured, and methods for completely restoring growth are currently unknown. All methods of therapy are aimed at reducing all negative manifestations of the disease to a minimum.
Causes
At the heart of this pathology are problems in the processes of bone formation due to a genetic failure in the formation of epiphyseal cartilage. The cells of the growth zone are arranged randomly, which provokes a disruption of normal ossification, and the patient’s growth slows down.
Only those bones that grow according to the enchondral type take part in this process. Since the bones located in the area of the cranial vault are formed from connective tissue, they grow according to age and this provokes an imbalance in proportion and the formation of a specific shape of the patient’s skull.
Symptoms
Anatomical abnormalities become noticeable after the baby is born. Doctors note such external manifestations in the newborn as:
- large volume head;
- limbs are very short.
It is clearly visible on the baby’s head that the forehead has a convex shape, and the occipital and parietal protuberances are protruding. In some cases, hydrocephalus is diagnosed. The eyeballs are set deep and the distance between them is wide. The shape of the nose is saddle-shaped, and the upper part is wide. The forehead, like the upper jaw, has a protrusion forward.
The arms and legs of patients with achondroplasia are shortened at the expense of the hips and shoulders. Doctors notice that a newborn’s hands can only reach the navel. The baby's body has normal development, there are no changes in the chest, the stomach is protruded forward.
Children diagnosed with achondroplasia are more likely than other children to experience sudden death in their sleep. Doctors explain this by saying that in such patients the medulla oblongata and the upper part of the spinal cord are compressed due to the small diameter of the hole in the occipital region.
In the first two years of life, a child develops kyphosis in the cervicothoracic region, which may disappear after the child begins to walk. All children with achondroplasia have delays in physical development, but mental and intellectual development does not suffer.
As the patient gets older, flatfoot feet begin to form, and the knee joints lose their stability. Adult women reach a height of 124 centimeters, and men - 131 centimeters. It is worth noting that most patients with this diagnosis tend to gain excess weight and suffer from obesity.
Diagnostics
Since the patient’s appearance has characteristic features, it is not difficult for doctors to make a diagnosis. All children should be examined to assess the severity of the pathology, and the data obtained should be recorded in a summary table. It is important to fill it out regularly, and the results are compared with the norm data developed for patients with achondroplasia.
It is also important to conduct research on different organs. For this purpose the following specialists are involved:
- neurosurgeon (he recommends MRI and CT);
- otolaryngologist;
- pulmonologist
Each patient is required to bring the doctor X-ray results, from which the doctor can confirm or refute the diagnosis.
On an x-ray of a patient with a history of achondroplasia, a violation of the proportions between such parts of the skull as the facial and cerebral parts is determined. The hole in the back of the head has a reduced diameter, and the size of the lower jaw and bones of the cranial vault is larger than in healthy patients.
A chest x-ray shows that the sternum is curved and the ribs are thickened and deformed. In some cases, the absence of the normal anatomical bend of the clavicle is determined.
The patient's spine does not show any special changes in the image, but its physiological curves are weakly expressed, which may lead to the development of lumbar hyperlordosis.
An X-ray of the pelvis shows a changed shape of the iliac wing. It is rectangular, unfolded and significantly shortened.
By taking X-rays of the joint, doctors can determine fibular lengthening, deformity, and incongruence.
Treatment
To date, there are no methods for completely curing patients from achondroplasia in orthopedics. Clinical studies have been conducted on the use of growth hormone, but this method has not proven its effectiveness.
- use special orthopedic shoes;
- engage in physical therapy;
- take a massage course;
- lose weight.
If the patient's arms and legs are very deformed, and the spinal cord canal is narrowed, then he is recommended to undergo surgery. In order to correct the deformity, doctors perform an osteotomy, and, if necessary, to eliminate the narrowing, a laminectomy.
In the case of leg lengthening to increase height, surgical intervention is performed crosswise in several stages. This means that in the first stage the thigh of the right leg and the shin of the left are operated on, and in the second - vice versa.
During the entire course of operations, the child can grow up to 28 centimeters. Therapy begins at 4 years of age. During this period, the first three stages of the intervention are carried out. Rehabilitation after each operation is about 5 months, and the interval between manipulations should not be less than 2-3 months.
For repeated stages, the child is invited at the age of 14–15 years. During this period, the teenager must again go through all stages of painful procedures.
Increase in height (video)