Solving problems in general and molecular genetics. Linked inheritance and crossing over Classic hemophilia and color blindness are inherited as recessive
58. In humans, the Rh factor locus is linked to the locus that determines the shape of red blood cells and is located at a distance of 4 morganids from it (K. Stern, 1965). Rh positivity and elliptocytosis are determined by dominant autosomal genes. One of the spouses is heterozygous for both traits. Moreover, he inherited Rh positivity from one parent, and elliptocytosis from the other. The second spouse is Rh negative and has normal red blood cells. Determine the percentages of probable genotypes and phenotypes of children in this family.
59. Nail-patella defect syndrome is determined by a completely dominant autosomal gene. At a distance of 10 morganids from it (K. Stern, 1965) there is a locus of blood groups according to the ABO system. One of the spouses has blood group II, the other - III. The one with blood type II suffers from defects in the nails and patella. It is known that his father was with blood group I and did not have these anomalies, and his mother, with blood group IV, had both defects. A spouse with blood type III is normal in relation to the gene for the defect of nails and patella and is homozygous for both pairs of analyzed genes. Determine the probability of birth in this family of children suffering from defects of the nails and patella, and their possible blood groups.
60. In the Drosophila fly, the gene for normal eye color dominates over the gene for white eyes, and the gene for abnormal abdominal structure dominates over the gene for normal structure. Both pairs of genes are located on the X chromosome at a distance of 4 morganids. Determine the probable genotypes and phenotypes of the offspring from crossing a female heterozygous for both traits with a male having normal eye color and normal abdominal structure.
61. Classic hemophilia and color blindness are inherited as recessive traits linked to the X chromosome. The distance between genes is determined to be 10 morganids.
a) A girl whose father suffers from both hemophilia and color blindness, and whose mother is healthy and comes from a family free from these diseases, marries a healthy man. Determine the likely phenotypes of the children from this marriage.
b) A woman whose mother suffered from color blindness and whose father suffered from hemophilia marries a man who suffers from both diseases. Determine the probability of children being born in this family simultaneously with both anomalies.
62. The gene for color blindness and the gene for night blindness, inherited through the X chromosome, are located at a distance of 50 morganids from each other (K. Stern, 1965). Both traits are recessive.
a) Determine the probability of having children simultaneously with both anomalies in a family where the wife has normal vision, but her mother suffered from night blindness, and her father suffered from color blindness, her husband is normal in relation to both signs;
b) Determine the probability of having children simultaneously with both anomalies in a family where the wife is heterozygous for both characteristics and inherited both anomalies from her father, and the husband has both forms of blindness.
63. In humans, the gene that determines the syndrome of nail and patella defect and the gene that determines the blood group in the ABO system are linked to each other and are located at a distance of 10 morganids. The gene that determines the Rh factor and the elliptocytosis gene are located on another chromosome and are located 4 morganids apart. Nail defect syndrome, elliptocytosis and Rh positivity are inherited in a dominant manner.
a) One spouse is heterozygous for all characteristics and has blood group IV. It is known that in previous generations no crossing over was observed in anyone, and he inherited the nail defect syndrome from his father along with the blood group II gene. The second spouse is homozygous for all recessive genes and has blood type I. Determine the likely phenotypes of the offspring from this marriage.
b) One spouse is heterozygous for nail defect syndrome, has blood group IV, Rh negative, red blood cells of normal shape. It is known that his father did not suffer from a nail defect and had blood type III. The second spouse has a normal nail structure and blood group I, and is heterozygous for the Rh factor and elliptocytosis. His mother was Rh positive and suffered from elliptocytosis. Calculate the likely phenotypes of the children in this family.
64. In humans, the genes for the Rh factor and elliptocytosis are located on the same chromosome at a distance of 4 morganids. Rh positivity and elliptocytosis are determined by dominant genes. The color blindness gene and the night blindness gene are located on the X chromosome at a distance of 50 morganids. Both traits are transmitted recessively.
a) A woman who is heterozygous for all characteristics, and whose ancestors did not have crossing over, marries a man who suffers from both color and night blindness and is homozygous for both autosomal recessive genes. Determine the likely phenotypes of the children in this family.
b) An Rh-positive woman with normal red blood cell shape and normal vision marries a Rh-negative man with elliptocytosis and night blindness. It is known that the woman’s father was Rh-negative and color blind, and her mother saw colors normally, but suffered from night blindness. Only the man's father suffered from elliptocytosis, and his mother also suffered from night blindness. Determine the probability of birth in this family of Rh-negative children without other anomalies.
c) A woman who is heterozygous for all characteristics marries an Rh-negative man who is normal in all other analyzed characteristics. It is known that the woman’s father was Rh-negative, had elliptocytosis, suffered from night blindness, but could distinguish colors normally. Determine the likely phenotypes of the children in this family.
65. It is known that “three-haired” cats are always female. This is due to the fact that the genes for black and red coat colors are allelic and are located on the X chromosome, but none of them dominates, and when red and black colors are combined, “three-coated” individuals are formed.
a) What is the probability of getting “three-haired” kittens in the offspring from crossing a “three-haired” cat with a black cat?
b) What kind of offspring can be expected from crossing a black cat with a red cat?
66. Enamel hypoplasia 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. What will their second son be like?
67. Classic hemophilia is transmitted as a recessive trait linked to the X chromosome.
a) A man with hemophilia marries a woman who does not have this disease. They give birth to normal daughters and sons, who marry people who do not suffer from hemophilia. Will hemophilia be discovered again in grandchildren and what is the likelihood of patients appearing in the families of daughters and sons?
b) A man with hemophilia marries a normal woman whose father suffered from hemophilia. Determine the probability of having healthy children in this family.
68. In humans, the gene that causes one of the forms of color blindness or color blindness is localized on the X chromosome. The disease state is caused by a recessive gene, the health state is caused by a dominant gene.
a) A girl with normal vision, whose father was color blind, marries a normal man, whose father was also color blind. What kind of vision can be expected in children from this marriage?
b) A man and woman who are normal in terms of vision have: a) a son who suffers from color blindness and has a normal daughter; b) a normal daughter with one normal son and one color-blind son; c) another normal daughter with five normal sons. What are the genotypes of parents, children and grandchildren?
69. Anhidrotic ectodermal dysplasia in humans is transmitted as an X-linked recessive trait.
a) A young man who does not suffer from this defect marries a girl whose father is deprived of sweat glands, and whose mother and her ancestors are healthy. What is the likelihood that the children from this marriage will suffer from lack of sweat glands?
b) A normal woman marries a man with anhidrotic ectodermal dysplasia. They have a sick daughter and a healthy son. Determine the probability of the next child being born without an anomaly.
70. Hypertrichosis is inherited as a trait linked to the Y chromosome. What is the probability of having children with this anomaly in a family where the father has hypertrichosis?
71. Darkening of teeth can be determined by two dominant genes, one of which is located on the autosomes, the other on the X chromosome. A family of parents with dark teeth gave birth to a daughter and a son with normal teeth color. Determine the probability of the birth of the next child in this family, also without anomalies, if it was possible to establish that the mother’s dark teeth are caused only by a gene linked to the X chromosome, and the father’s dark teeth are caused by an autosomal gene for which he is heterozygous.
72. One form of agammaglobulinemia is inherited as an autosomal recessive trait, the other as an X-linked recessive trait. Determine the probability of having sick children in a family where it is known that the mother is heterozygous for both pairs of genes, and the father is healthy and has only dominant genes of the analyzed alleles.
73. In humans, color blindness is caused by an X-linked recessive gene. Thalassemia is inherited as an autosomal dominant trait and is observed in two forms: in homozygotes it is severe, often fatal, in heterozygotes it is less severe. A woman with normal vision but mild thalassemia, married to a healthy but colorblind man, has a colorblind son with mild thalassemia. What is the probability of having the next son without anomalies?
74. In humans, classical hemophilia is inherited as an X-linked recessive trait. Albinism is caused by an autosomal recessive gene. One married couple, normal according to these two characteristics, had a son with both anomalies. What is the probability that the second son in this family will also develop both anomalies at the same time?
75. A man suffering from color blindness and deafness married a woman with normal vision and good hearing. They had a son who was deaf and colorblind and a daughter who was colorblind but had good hearing. Determine the probability of having a daughter in this family with both anomalies, if it is known that color blindness and deafness are transmitted as recessive traits, but color blindness is linked to the X chromosome, and deafness is an autosomal trait.
76. Hypertrichosis is transmitted through the Y chromosome, and polydactyly is transmitted as a dominant autosomal trait. In a family where the father had hypertrichosis and the mother had polydactyly, a daughter was born normal in terms of both characteristics. What is the probability that the next child in this family will also be without both anomalies?
77. Hypertrichosis is inherited as a Y-chromosome-linked trait that does not appear until the age of 17. One form of ichthyosis is inherited as a recessive trait linked to the X chromosome. In a family where the woman is normal in both signs, and the husband has only hypertrichosis, a boy was born with signs of ichthyosis.
a) Determine the probability of hypertrichosis in this boy.
b) Determine the probability of having children in this family without both anomalies, and what gender they will be.
78. Retinitis pigmentosa can be inherited in three ways: as an autosomal dominant, autosomal recessive and X-linked recessive trait. Determine the probability of having sick children in a family where the mother has retinitis pigmentosa and is heterozygous for all three pairs of genes, and the father is healthy and homozygous for all three pairs of genes.
79. A right-handed woman with brown eyes and normal vision marries a right-handed man with blue eyes and color blindness. They had a blue-eyed daughter, left-handed and colorblind. What is the probability that the next child in this family will be left-handed and suffer from color blindness, if it is known that brown eye color and the ability to use predominantly the right hand are dominant, autosomal, unlinked traits, and color blindness is a recessive trait linked to the X chromosome ? What eye color is possible in sick children?
80. In a family where the wife has blood group I and the husband IV, a color-blind son was born with III group blood. Both parents see colors normally. Determine the probability of having a healthy son and his possible blood types. Color blindness is inherited as a recessive trait linked to the X chromosome.
81. Parents with blood type II had a son with blood type I and a hemophiliac. Both parents do not suffer from this disease. Determine the probability of a second child being born healthy and its possible blood types. Hemophilia is inherited as a recessive trait linked to the X chromosome.
82. Otosclerosis is inherited as a dominant autosomal trait with a penetrance of 30%. The absence of upper lateral incisors is inherited as an X-linked recessive trait with full penetrance. Determine the probability of children exhibiting both anomalies simultaneously in a family where the mother is heterozygous for both traits and the father is normal for both pairs of genes.
83. John Spock, second in command of the transgalactic starship Enterprise, was the son of two planets: Vulcania and Earth. His father was Vulcan and had pointed ears (P), a right-sided heart (R), and underdeveloped adrenal glands (A), all traits that are dominant to Earth alleles. His mother was a dugout. Genes A and R are located on the same chromosome at a distance of 20 map units. The P gene lies in a different linkage group. If John marries Earth's daughter, what is the probability that:
a) Will the child have the same set of characteristics as his father?
b) The first child will have underdeveloped adrenal glands, and the remaining two signs will be earthly?
c) That they will have two children: one, by all indications, an Earthling, and the other a Vulcan?
Please note that the penetrance of the P gene is 90%, R is 85%, and the A gene is 100%.
6.2. Answer security questions.
7. Questions for tests on this topic:
7.1. .Questions for current computer controls:
1. Crossing over occurs:
A in interphase I, in period S;
b in prophase I, at the pachytene stage;
V in prophase I, at the zygotene stage;
G in anaphase I of division;
d in anaphase II division.
2. With partially linked inheritance, if the distance between genes is more than 50 morganids, in a diheterozygous organism (AaBb) the following is formed:
A two types of gametes;
b two types of crossover (20% each) and two types of non-crossover gametes (40% each);
V one type of gamete;
G only crossover gametes;
d four types of gametes are formed in equal shares of 25%.
3. The position of the chromosomal theory of heredity:
A genes are located on the chromosome in a linear order and are inherited together;
b independent combination of traits in F2 hybrids;
V hybrids exhibit uniformity in F1;
G genes located on the Y chromosome are passed on from father to son;
d the splitting of characters in hybrids in F2 occurs in a ratio of 9:3:3:1.
4. In the karyotype of men, the X and Y chromosomes are singular, so all genes (dominant and recessive) will be implemented into a trait. This phenomenon is called:
A the phenomenon of hemizygosity;
b the phenomenon of homogamety;
V the phenomenon of heterogamety;
G holandric;
d recognition.
5. Determine the probability of hypertrichosis in children in a family where the mother is healthy and the father is sick; it is known that all male relatives were sick:
A 100% of all children;
b 100% among boys;
V 50% among boys, 25% among girls;
G 50% among boys, all girls are healthy;
d all children are healthy.
6. With complete linkage of genes in a diheterozygous organism, AaBb is formed:
A two types of gametes;
b two types of crossover gametes;
V two types of non-crossover and two types of crossover gametes;
G one type of gamete;
d only crossover gametes.
7. The number of clutch groups corresponds to:
A diploid number of chromosomes;
b haploid number of chromosomes;
V crossover probabilities;
G a value less by one than the haploid number of chromosomes (n-1);
d none of the above answers.
8. Holandric inheritance is the inheritance of traits whose genes are localized:
A in homologous regions of the X chromosome;
b in autosomes;
V on the Y chromosome;
G in non-homologous regions of the X chromosome;
d on chromosome 1 of the human karyotype.
7.2. Questions included in the course exam tickets:
1. Characteristics of Drosophila as a genetic object.
2. Chromosome mapping methods.
3. The phenomenon of gene linkage. Clutch groups and their number. Crossing over. Probability of crossing over. Chromosomal theory of heredity.
4. Inheritance of sex-linked characteristics:
· list the loci of complete and partial linkage to the X chromosome;
· holandric traits and the nature of their inheritance;
5. Traits limited by sex and controlled by sex. Definition. Examples.
8. Literature to prepare for the lesson:
8.1. Main:
8.1.1. Biology, textbook UMO, 2t., ed. Yarygina V.N., Mir, 2002
8.1.2. Biology with genetics, textbook UMO, 2 volumes, ed. Yarygina V.N., Mir, 2000.
8.2. Additional:
8.2.1. Biology. I.V. Chebyshev, G.G. Grineva, M.V. Mower. M.: GOU VUNMC, 2001.
8.2.2. Cytology. Educational and methodological manual for students. Ekaterinburg, 2010
8.2.3. Pekhov A.P. Biology with basics of ecology. Series “Textbooks for universities. Special literature" - St. Petersburg: Lan Publishing House, 2000.
8.2.4. General and medical genetics: A textbook for students of higher medical educational institutions. /V.P. Shchipkov, G.N. Krivoshein. – M.: Publishing Center “Academy”, 2003.
8.2.5. Vogel F., Motulski A. Human genetics (in 3 volumes). M., "Mir", 1989.
8.2.6. Shevchenko V.A. and others. Human genetics. Textbook for students of higher educational institutions. M.: Humanitarian publication center VLADOS, 2002.
GBOU VPO "Ural State Medical Academy"
Ministry of Health and social development
I affirm:
Head department of medical
Biology and genetics
Doctor of Medical Sciences, Prof. Makeev O.G.
________________________
"___"_____________2011
Study assignment
for students
Faculty: therapeutic and prophylactic
Course 1 Semester 1
Lesson No. 10
Lesson topic: Fundamentals of medical genetics.
Genes located in the same linkage group are called linked, i.e. in one pair of homologous chromosomes. A woman has 24 linkage groups (23+M chromosome), a man has 25, becauseY -chromosome is a separate linkage group.
Task No. 1
In humans, the development of color blindness and hemophilia is caused by the action of recessive genes localized on the X chromosome at a distance of 9.8% centimorganids from each other.
Healthy woman, whose father suffered from hemophilia, marries a healthy man. It is known that the woman’s mother was color blind, but there was no hemophilia among her relatives.
1. What percentage of a woman’s gametes contain both hemophilia and color blindness genes?
2. How many different genotypes can there be among the children of this married couple?
Solution to problem No. 1:
X D H - does not carry the genes for hemophilia and color blindness.
X d h - carries the genes for hemophilia and color blindness.
P ♀ X D h X d H x X D H Y ♂
G X D h ; X d H by 45.1% X D H ;Y
X D H , X d h 4.9% each
F1 X D H X D h ;
X D h Y;
X D H Y;
X D H X d H ;
3. Nail and patella defect syndrome is determined by a completely dominant autosomal gene. At a distance of 10 centimorganids from it there is a locus of blood groups according to the ABO system. One of the spouses has blood group II, the other – III. Anyone with blood group II suffers from defects in the nails and kneecap. It is known that his father had blood group I and did not have these anomalies, and his mother had blood group IV and both defects. A spouse with blood group III is normal in relation to the gene for the defect of nails and patella and is homozygous for both pairs of analyzed genes.
Determine the probability of the birth of children in this family suffering from defects of the nails and patella and their possible blood groups.
4. Cataracts and polydactyly in humans are caused by dominant autosomal closely linked (i.e., not detecting crossing over) genes. However, not necessarily the genes for these anomalies may be linked, but also the cataract gene with the gene for normal hand structure and vice versa. The woman inherited cataracts from her mother and polydactyly from her father. Her husband is normal for both signs. What is more likely to be expected in their children: the simultaneous appearance of cataracts and polydactyly, the absence of both of these signs, or the presence of only one anomaly - cataracts or polydactyly?
5. Cataracts and polydactyly in humans are caused by dominant autosomal closely linked (i.e., not detecting crossing over) genes. However, not necessarily the genes for these anomalies may be linked, but also the cataract gene with the gene for normal hand structure and vice versa. The woman inherited cataracts from her mother and polydactyly from her father. Her husband is normal for both signs.
a) What kind of offspring can be expected in a family where the husband is normal and the wife is heterozygous for both characteristics, if it is known that the wife’s mother also suffered from both anomalies and her father was normal?
b) What kind of offspring can be expected in a family from parents who are heterozygous for both traits, if it is known that the mothers of both spouses suffered only from cataracts, and the fathers only from polydactyly?
6. Classic hemophilia and color blindness are inherited as recessive traits linked to the X chromosome. The distance between genes is determined to be 9.8 centimeters. A girl whose father suffers from both hemophilia and color blindness, and whose mother is healthy and comes from a family free from these diseases, marries a healthy man. Determine the likely phenotypes of the children from this marriage.
7. The gene for color blindness (color blindness) and the gene for night blindness, which are inherited through the X chromosome, are located at a distance of 50 centimorganids from each other. Both traits are recessive. Determine the probability of having children simultaneously with both anomalies in a family where the wife has normal vision, but her mother suffered from night blindness, and her father suffered from color blindness, while the husband is normal in relation to both signs.
Any science has its own specific vocabulary, which is quite difficult for an uninitiated person to understand. A clear confirmation of this is genetics - one of the most complex branches of biology. Many people have an idea that genes are present in the sex chromosomes of the human body. They not only determine the gender characteristics of women and men, but also influence other features of the functioning of the body. Scientists were able to find out this pattern when studying a number of diseases. In particular, it was found that rare diseases such as hemophilia and color blindness are recessive traits that carry certain genetic information. In this case, the recessive gene is linked to the X chromosome. Before explaining the essence of this process, it is necessary to make a small clarification. Male cells contain one Y chromosome, which determines the sex of the male body and is responsible for the formation of sperm, and one X chromosome. Female cells contain only two (sometimes three) X chromosomes.
Hemophilia
This disease, which predominantly affects men, is a very dangerous pathology. In hemophilia, a person's blood loses its ability to clot. Even a minor wound or a small cut in such people can lead to prolonged bleeding. A large loss of blood is a mortal threat to human life. The problem lies in the X chromosome. The gene it contains can be dominant, determining blood clotting, or recessive, preventing blood from clotting normally.
Some women have a recessive gene on one of their X chromosomes, but they do not even suspect that they are its carrier. On the other X chromosome, the dominant gene works as expected, so the blood clots normally. But if a defective gene is found in a man, it is dangerous. After all, men only have one X chromosome. At the same time, a man can receive a recessive gene exclusively from his mother; accordingly, a son cannot inherit hemophilia from his father. In general, a woman with hemophilia is an extremely rare phenomenon.
As an example, it is worth mentioning the hemophiliac Tsarevich Alexei, the son of Nicholas II, the last Russian Tsar. He inherited the recessive gene from his mother. It is assumed that Empress Alexandra Feodorovna received the hemophilia gene from her grandmother, Queen Victoria of England, who, according to historians, was its carrier.
Colorblindness
Another well-known disease is color blindness - color blindness. The development of this pathology follows a similar scenario and is also associated with the presence of a defective gene on the X chromosome. Therefore, in this case, only women are carriers of the recessive gene. A non-colorblind man and a carrier woman have an equal chance of giving birth to both a colorblind son and a healthy boy.
In the opposite situation (healthy mother and colorblind father), only their daughter can inherit the recessive gene, and there is no need to worry about the health of the son. And the highest probability of giving birth to a daughter with abnormal color perception (more than 50%) occurs when the parents of the unborn child are a couple of a color-blind man and a carrier woman. Indeed, in this case, along with the maternal gene for color blindness, the girl additionally receives a defective paternal X chromosome.
Genes located on sex chromosomes determine traits that are called sex-linked. This applies not only to hemophilia and color blindness. In particular, as many as 267 traits are transmitted to humans through the X chromosome. An example would be rickets, which cannot be cured with vitamin D, or the birth of a person with brown tooth enamel.
Of course, genetics is the science of the future. Perhaps someday people will learn to penetrate the human hereditary apparatus and will be able to protect it from negative impact environment and even eliminate or replace defective genes. But for this it is necessary to solve many not only scientific, but also moral and ethical problems.
non-crossover 100-10=90% number of non-crossover gametes, each gamete accounts for 90:2=45%, 45x1/2=22.5%.
crossover The number of crossover gametes is 10%, for each gamete there are 10:2=5%, 5x1/2=2.5%
Healthy boys develop from a zygote formed by combining an egg with normal genes from the non-crossover part of gametes with a sperm cell
Answer: the probability of having healthy boys is 22.5%.
In many organisms, the Y-sex chromosome, unlike the X-sex chromosome, is hereditarily inert, since it does not contain genes; its function is to determine sex, so in humans, genes located on the Y-sex chromosome determine the male sex. In this regard, we do not designate genes on the Y sex chromosome with the exception of the hypertrichosis gene and the gene that causes the film between the II and III toes.
Problem 3
To determine the genotype and phenotype of children based on the location of genes in sex and autosomal chromosomes
Hypertrichosis (hair growth on the edge of the auricle and earlobe during puberty in young men) is transmitted through the Y chromosome, and polydactyly (six-fingered) is an autosomal dominant trait. In a family where the father had hypertrichosis and the mother had polydactyly, a daughter was born normal in relation to both characteristics. What is the probability that the next child will not have both anomalies?
Hypertrichosis is inherited holandically, i.e. with the Y-sex chromosome, it manifests itself only in the male line, therefore all men born in that marriage will have the hypertrichosis gene, which they receive from their father through the Y-sex chromosome. This gene appears during puberty in young men. The father did not suffer from polydactyly, but had hypertrichosis, his genotype was aaXY 1. The mother had six fingers, but in this family a daughter was born with a normal number of fingers, therefore the woman was heterozygous for this gene and gave her daughter a normal gene, her AaXX genotype.
Let us designate the genes and write down the marriage pattern:
Solution
F: ♀ aaXX - healthy daughter
Define F - healthy?
In this marriage, all sons born at puberty will have hypertrichosis, half of the female daughters, 1/4 of the total offspring will have polydactyly, 1/4 of the sons of the total offspring and 50% of the male offspring will also suffer from polydactyly, i.e. half of all offspring, regardless of gender, will inherit the autosomal dominant polydactyly gene, and for both characteristics the number of affected children is 75% of all offspring.
Only daughters of 1/4 (25%) of all offspring, and 1/2 . (50%) from the female sex, i.e. they will have the normal gene for developing five fingers.
Answer: all girls will be healthy with respect to hypertrichosis, but half of the daughters will receive the abnormal polydactyly gene - this will amount to 50% of the female offspring.
Generally 1 /4 children (daughters only), 25% of all offspring will be healthy.
III. 12 Problems on sex-linked inheritance 1. The absence of sweat glands in humans is inherited as a recessive trait linked to the X chromosome. A young man who does not suffer from this defect marries a girl whose father lacks sweat glands, and whose mother and her ancestors are healthy. What is the probability that the sons and daughters of this marriage will lack sweat glands?
2. In the laboratory, red-eyed fruit flies were crossed with red-eyed males. The offspring included 69 red-eyed and white-eyed males and 71 red-eyed females. Write the genotypes of the parents and offspring if it is known that red eye color is dominant over white, and the genes for eye color are located on the X chromosome.
3. The daughter of a hemophiliac marries the son of a hemophiliac. Hemophilia is inherited recessively linked to the X chromosome. Determine the probability of having healthy children in this family?
4. The body color of cats and the gooseberry moth is controlled by a sex-linked gene localized on the X chromosome. In two experimental crosses in which the homogametic sex in the parent generation was homozygous for body color, the following results were obtained:
MothCat
(normal color is dominant - (black color is dominant
flies over the pale one) over the redhead)
rhodi phenotypes - pale male x normal black male x red
female female
Ratio 1 normal male 1 red male
phenotypes in each - 1 normal female 1 black female
Which sex is heterozygous for each of these organisms?
5. Some chicken breeds have genes that determine White color and striped plumage coloring, linked to the X chromosome, striping dominates over the white solid color. The heterogametic sex in chickens is female. On a poultry farm, white chickens were crossed with striped roosters and got striped feathers in both roosters and chickens. Then the individuals obtained from the first crossing were crossed with each other and 594 striped roosters and 607 striped and white chickens were obtained. Determine the genotypes of parents and descendants of the first and second generations.
6. What children can be born from the marriage of a hemophiliac with a woman suffering from color blindness (but otherwise having a completely favorable genotype)?
7. A man suffering from color blindness and deafness married a woman with normal vision and good hearing. They had a son who was deaf and colorblind and a daughter who was colorblind but with good hearing. Determine the probability of having a daughter in this family with both anomalies, if it is known that color blindness and deafness are transmitted as recessive traits, but color blindness is linked to the X chromosome, and deafness is an autosomal trait.
8. It is known that “three-haired” cats are always female. This is due to the fact that the genes for black and red coat colors are allelic and are located on the X chromosome, but none of them dominates, and when red and black colors are combined, “three-coated” individuals are formed.
1) what is the probability of getting three-haired kittens in the offspring from crossing a three-haired cat with a black cat?
2) what kind of offspring can be expected from crossing a black cat with a red cat?
9. A right-handed woman with brown eyes and normal vision marries a right-handed, blue-eyed, colorblind man. They had a blue-eyed daughter, left-handed and colorblind. What is the probability that the next child in that family will be left-handed and suffer from color blindness, if it is known that brown eye color and the ability to use predominantly the right hand are dominant autosomal unlinked traits, and color blindness is a recessive trait linked to the X chromosome? What eye color is possible in sick children?
10. Darkening of teeth can be determined by two dominant genes, one of which is located on the autosome and the other on the X chromosome. A family of parents with dark teeth gave birth to a daughter and a son with normal teeth color. Determine the probability of the birth of the next child in this family, also without anomalies, if it was possible to establish that the mother’s dark teeth are caused only by a gene linked to the X chromosome, and the father’s dark teeth are caused by an autosomal gene for which he is heterozygous.
11. In humans, albinism is caused by an autosomal recessive gene. Anhidrotic ectodermal dysplasia is transmitted as an X-linked recessive trait. A married couple, normal in both characteristics, had a son with both anomalies. What is the probability that their second child will be a girl normal in both characteristics?
12. In a family where the wife has blood type I and the husband IV, a son was born - color blind with blood type III. Both parents can distinguish colors normally. Determine the probability of having a healthy son and his possible blood types. Color blindness is inherited as a recessive trait linked to the X chromosome.
13. The gene for color blindness (color blindness) and the gene for night blindness, inherited through the X chromosome, are located at a distance of 50 morganids from each other (K. Stern, 1965). Both traits are recessive. Determine the probability of children being born simultaneously with one and both anomalies in a family where the wife has normal vision, but her mother suffered from night blindness, and her father suffered from color blindness, while the husband is normal in relation to both signs.
14. In the Drosophila fly, the gene for normal eye color (red) dominates over the gene for white eyes, and the gene for normal abdominal structure dominates over the gene for its abnormal structure. This pair of genes is located on the X chromosome at a distance of 3 morganids. Determine the probability of different geno- and phenotypes in the offspring from crossing females heterozygous for both traits and males with normal eye color and normal abdominal structure.
15. Classic hemophilia and color blindness are inherited as recessive traits linked to the X chromosome, the distance between genes is determined at 9.8 morganids. A woman whose mother was color blind and whose father was hemophiliac marries a man who is color blind. Determine the probability of children being born in this family simultaneously with both anomalies.
16. A man suffering from color blindness and deafness, having a second blood group, married a woman with normal vision, good hearing, and a third blood group. They had a son who was deaf, colorblind, and had blood type O. What is the probability of having deaf children suffering from color blindness with blood type IV?
17. A brown-eyed woman with normal blood clotting and a third blood group, whose father had Blue eyes and suffered from hemophilia, marries a blue-eyed man who has normal blood clotting and a second blood group (heterozygous for this gene). What is the probability of having blue-eyed, hemophilic children with a second blood group?
18. In humans, a recessive gene causes color blindness, another recessive gene causes Duchenne muscular dystrophy. Both traits are inherited in a sex-linked manner. According to the pedigree of one large family The following data were obtained: a healthy woman with normal vision, whose father suffered from muscular dystrophy and whose mother suffered from impaired color vision, married a healthy man with normal color vision. From this marriage 8 boys and 3 girls were born. Of these: 3 girls and 1 boy were healthy and had normal vision. Of the remaining boys: 3 suffered from muscular dystrophy, 3 from color blindness and 1 from both diseases. Based on these data, an approximate (due to the paucity of material) estimate of the distance between the two named genes was given. Determine this distance?
19. Albinism is determined by an autosomal recessive trait, hyperplasia (darkening of tooth enamel) by a dominant gene linked to the X-sex chromosome. A woman with dark teeth, normal skin color and third blood group marries a man with normal teeth color, normal skin pigmentation and first blood group. They had a son with normal teeth color, an albino and with the first blood group. What is the probability of having the next son with normal teeth color, an albino and a third blood type?
Ш.13 Determination of gametes with independent inheritance, complete and incomplete linkage of genes.Independent inheritance
The number of gamete types with independent inheritance is determined by the formula 2 n, where n is the degree of heterozygosity. (See chapter di - and polyhybrid crossing).
Determine the probable types of gametes for the genotype AaBBCcDd, the genes of which are located in different pairs of chromosomes. This genotype is triheterozygous for genes A, C, D and homozygous for the recessive gene B. The number of gamete types for the specified genotype 2 = 8, because genes from different pairs can be freely combined. The genes for a homozygous BB combination will be equally present in each gamete.
Solution: Let's write down the location of the genes of the AaBBCcDd genotype cytologically, i.e. in different pairs of chromosomes
The gamete must include a haploid set containing one gene from each of the three pairs of allelic genes. Let us first write down the types of gametes containing the dominant gene A:
1) AbCD; 2) AbCd; 3) ABCD; 4) Abcd
the same number of gamete types will be recessive
1) abcd; 2) abcD; 3) abCd; 4) abCD
The number of gamete types, 8, can be divided by 2, because half of the gametes carry the dominant allele A, and half carry the recessive allele a.
When determining the four types of gametes with the recessive gene - a, we record combinations of genes in gametes alternatively in comparison with gametes having the dominant gene A, due to independent chromosome divergence and, accordingly, independent combination of genes.
Answer: genotype AaBBCcDd produces 8 types of gametes, 12.5% each
II incomplete gene linkage
We draw the location of genes in a homologous pair of chromosomes. Since the linkage of genes is incomplete, non-crossover and crossover gametes will be formed. The number of non-crossover gametes is greater than crossover ones if the distance between genes in the linkage group is less than 50%.
In this problem, the percentage of crossing over is not specified, so we assume the condition that the distance between genes is less than 50%.
There are 2 non-crossover gametes, both with complete linkage and with incomplete linkage.
As a result, 6 types of crossover gametes are formed:
Since the genotype is heterozygous for three genes M, P and E, and the linkage of genes is incomplete, three crossings are possible. We change allelic genes, then
III Genes are located V different pairs of chromosomes
According to the third condition of the problem, when genes are located in different pairs of chromosomes, therefore they are not linked, the number of gametes is determined by the formula 2 n, where n is the degree of heterozygosity. In this case, the Drosophila fly is heterozygous for three genes M P and E, so the number of gametes is 2 3 = 8, in equal proportions.
We draw the location of genes in different pairs of chromosomes. 
Independent divergence of chromosomes in anaphase I division of meiosis produces 8 combinations.
8 types of gametes 12.5% each: (100%:8=12.5)
Answer: 1) with complete linkage of genes, 2 types of gametes are formed, 50% each MORSE And torso(gametes only non-crossover);
2) In case of incomplete linkage of genes, a total number of gametes is 8, among which 2 gametes are non-crossover MORSE And torso and 6 types of crossover gametes: TORSO, Morse, MorsE, torso, MORse, torso. The total number of non-crossover gametes is greater than crossover gametes;
3) When genes are located in different pairs of chromosomes, 8 types of gametes are formed, 12.5% each
MORSE and TORSE; MORSE and TORSE; Morse and Torse; Morse and Torse
III. 14 Tasks for the formation of gametes with independent inheritance, incomplete and complete linkage of genes
1. In corn, smooth seeds (S) dominate over wrinkled ones (s) and colored ones (C) dominate over uncolored ones (c). The S and C genes are localized in one homologous chromosome, the s and c genes are located in another homologous chromosome at a distance of 3 morganids. Determine: what types of gametes and in what quantitative ratio are formed in plants with the SsCc genotype and the following gene linkage:
2. Drosophila has a recessive gene - b, which determines the black body, and a gene - p | determining purple eye color, and, accordingly, the dominant genes B (gray color), P (red eyes) are linked and located in the second autosomal pair of homologous chromosomes at a distance of 6 morganids. Determine: what types of gametes and in what ratio are formed in individuals with genotype BvPp I and the following gene linkage:
3. Genes A, a and B, b are located in one homologous pair of autosomal chromosomes and are completely linked. Determine: what types of gametes and in what quantities are formed in individuals with the AaBb genotype and the following gene linkage:
4. Genes C, c and D, d are localized in one pair of homologous chromosomes and are absolutely linked, genes M, m and N, n are localized in another pair of homologous chromosomes at a distance of 12 morganids. Determine what types of gametes and in what quantities are formed in individuals with the genotype CcDdMmNn and the following gene linkage:
5. How many types of gametes and in what ratio is an individual with the genotype AaBBCcDdEe formed, if all genes are linked and localized in one homologous pair of autosomal chromosomes, with complete linkage of genes L and B and D and e, incomplete linkage of genes Vis, the distance between which is 16 morganids .
6. What types of gametes and in what ratio form genotypes with independent inheritance of autosomal genes:
1)ВВСs; 2)ВвСс; 3)BвCcDDEe; 4) BвCcDdЕe; 5) AABBCCDDEE; 6) MMNnPPooKKFF.
7. What types of gametes and in what quantitative ratio does an individual with the genotype AaBBCcDdFfHhGg form, given the linked localization of the ABcdF genes in one homologous autosomal chromosome, and the abCDf genes in another. The BcdF and BCDf genes are completely linked, the distance between genes A and B and a and b corresponds to 8 morganids. The H and G genes and, accordingly, the h and g genes are absolutely linked in another homologous pair of autosomal chromosomes.
8. What types of gametes and in what ratio does the AaBbCc genotype form when genes A and B and respectively a and B are incompletely linked to one homologous pair of chromosomes, the distance between the genes is 18 morganids, genes C and C are localized in another pair of homologous chromosomes.
9. According to some pedigrees in humans, the dominant gene for elliptocytosis (E, irregular shape of erythrocytes) and the gene that determines the presence of Rh antigen in erythrocytes (D) are localized in the first pair of homologous autosomal chromosomes at a distance of 20 morganids (according to other data, at a distance of 3- x morganid). What types of gametes and in what quantity are formed in a man with genotype 33Dd with the following gene linkage:
10. What types of gametes and in what quantitative ratio does an individual with the genotype CCDdttMmFf form when:
1) complete linkage of genes;
2) incomplete linkage of genes;
3) localization of genes in different pairs of chromosomes?
111.15 Problems for repetition of material
1. In the Drosophila fly, genes C, c and D, d are located in different pairs of homologous autosomes, genes P and r are linked and are located on the X-sex chromosome, respectively, genes p and R are in another homologous X-sex chromosome, the relative distance between the genes is 16 morganids. How many and what types of gametes are produced by a female diheterozygous for autosomal genes C and D and diheterozygous for genes P and R, located linked in the X-sex chromosomes?
2. Healthy parents gave birth to a son suffering from color blindness with Klinefelter syndrome. What are the genotypes of the parents and child? Explain the mechanism causing the karyotype disorder, determine the form of mutational variability, the total number of chromosomes?
3. The following karyotype 4A+ХХУУУ was established in human embryonic fibroblast cells. Explain the mechanism of chromosome set disorder. What are the consequences of such a mutation, the form of mutational variability, the total number of chromosomes?
4. According to a preliminary estimate, the relative distance between the genes for color blindness and Duchenne muscular dystrophy, located on the X sex chromosome and inherited recessively, is 25 morganids. Albinism (lack of melanin pigment) is inherited in an autosomal recessive manner. A healthy woman, who phenotypically received the genes for color blindness and albinism from her father, and the gene for Duchenne muscular dystrophy from her mother, marries an albino man who has normal genes that determine normal vision and the functioning of the muscular system. From this marriage an albino son was born, suffering from color blindness and muscular dystrophy. What is the prognosis for the birth of healthy offspring based on the phenotype in this marriage?
5. During mitosis in human tissue culture, elimination of one chromosome occurred. How many chromosomes will be in the two resulting cells?
6. If the original cell has 28 chromosomes, then how many chromosomes and chromatids go to one pole in anaphase of reduction division? How many chromatids go to each pole in anaphase of the equational (second meiotic) division?
7. When analyzing the crossing of a triheterozygote, the following splitting by phenotype was obtained: A-B-C- - 126, A-bb-C- - 120, aaB-C- - 128, aabbC-- 136, A-B-cc - 114, A-bbcc - 122, aaB-cc - 112, aabbсc - 126. What can be said about the localization of genes?
8. When black mice are crossed with each other, black offspring are always obtained. When yellow mice were crossed with each other, 111 black mice and 223 yellow mice were obtained. Which traits are dominant and which are recessive? What are the genotypes of parents and offspring? How can the mechanism of progeny splitting be explained?
9. It is known that “tricolored” cats are always female. This is due to the fact that the genes for black and red coat colors are allelic and are located on the X chromosome, but none of them dominates, and when black and red colors are combined, “three-coated” individuals are formed.
1) what kind of offspring can be expected from crossing a black cat with a red cat?
2) when crossing a three-haired cat with a ginger cat, a three-haired male kitten was noted in the litter. Explain the mechanism that caused this phenotype in the kitten, determine the form of mutational variability?
10 The red color of the onion bulb is determined by the dominant gene, yellow - by its recessive allele. However, the manifestation of the color gene is possible
only in the presence of another, unlinked dominant gene; the recessive allele of which suppresses color, the bulbs turn out to be white. Red bulbous plants were crossed with yellow bulbous plants and the resulting offspring were:
195 - red bulbous plants,
194 - yellow bulbous,
131 plants with white bulbs.
Determine the genotypes of parents and offspring, types of interaction of non-allelic dominant genes?
11. One of the fragments of the m-RNA molecule has the nucleotide sequence: AUG, AGC, GAC, UCG, ACC. As a result of the mutation, the seventh guanyl nucleotide was replaced by a cytidyl nucleotide. Determine how this will affect the polypeptide, which t-RNA anticodons will correspond to this segment of the mutational m-RNA.
12. In the Drosophila fly, gray body color dominates over black, normal wings over rudimentary wings. Drosophila flies with a gray body and normal wings were crossed, the offspring were Drosophila flies: with a gray body and normal wings, with a gray body and rudimentary wings, with a black body and normal wings and a black body with rudimentary wings in a ratio of 5:1:1:1 . Explain the nature of splitting, determine the genotypes of parents and offspring?
13 When pea plants with smooth seeds and red flower color were crossed with plants with wrinkled seeds and white flowers, in the first generation all plants had a smooth seed shape and. pink flowers. In the second generation it was obtained:
Smooth with red flowers - 246,
Pink with smooth seeds - 480,
White with smooth seeds - 242,
Red with wrinkled seeds - 86,
Pink with wrinkled seeds - 156,
White with wrinkled seeds - 80.
How are the traits being studied inherited? What are the genotypes of parents and offspring?
Appendix 2 Complete Domination
| HUMAN | |
| Dominant trait | Recessive trait |
| 1. Brown eyes | Blue or grey eyes |
| 2. Right-handedness | Left-handedness |
| 3. Dark hair | Blonde hair |
| 4. Normal pigmentation of skin, hair, eyes. | Albinism (lack of pigmentation) |
| 5. Big eyes | small eyes |
| 6. Thick lips | thin lips |
| 7. "Roman nose" | straight nose |
| 8. Gout | normal bone joints |
| 9. Normal carbohydrate metabolism | diabetes |
| 10. Polydactyly (extra fingers) | normal number of fingers |
| 11. Short-fingered (brachydactyly) | normal finger length |
| 12. Freckles on the face | no freckles |
| 13. Short body height | normal body growth |
| 14. Normal hearing | congenital deaf-muteness |
| PEAS | |
| 1. Yellow color of seeds | green color of seeds |
| 2. Smooth seed surface | wrinkled seed surface |
| 3. Red color of the corolla | white color of the corolla |
| 4. Axillary flower | terminal flower |
| 5. Tall | dwarf stature |
| FIGURED PUMPKIN | |
| 1. White color of the fruit | yellow color of the fruit |
| 2. Disc-shaped fetus | spherical (round) shape of the fruit |
| TOMATO | |
| 1. Spherical (round) shape of the fruit | pear-shaped fruit |
| 2. Red color of the fruit | yellow color of the fruit |
| 3. Tall stem | dwarf stem |
| 4. Purple stem | green stem |
| OVES | |
| 1. Early ripeness | late ripening |
| 2. Normal height | giant growth |
| 3. Immunity against rust | non-immune, susceptible to rust |
Inheritance of sex-linked traits.
Many traits in humans are inherited linked to sex chromosomes. Genes localized on sex chromosomes have their own characteristics of transmission over generations. For example, hemophilia, color blindness, etc. are inherited linked to the X chromosome, and the gene for baldness, hypertrichosis (hair growth in the ear), etc. are inherited with the Y chromosome. Since the X and Y chromosomes, along with homologous regions, also have non-homologous ones, When solving problems of such inheritance, it is necessary to take into account which of the sex chromosomes the corresponding gene is located on, the X or Y chromosome. It is more convenient to record with the image of sex chromosomes. For example, X h is a hemophilia gene linked to the X chromosome. There is no hemophilia gene on the Y chromosome. Therefore, possible genotypes of people for this trait may be as follows:
X H X h is a woman who is a carrier of the hemophilia gene, but is phenotypically healthy and has normal blood clotting.
X h Y – a man with hemophilia, because he does not have the allelic gene on the Y chromosome, then the recessive gene will manifest itself phenotypically.
X h X h – female hemophiliac, very rare, because die early or become infertile.
Analysis of the results of crossing with linked inheritance is carried out separately for the female and male sexes.
Tasks
49. A girl whose father is a hemophiliac marries a healthy man. What kind of children can be expected from this marriage?
50. Classic hemophilia is a recessive trait linked to the X chromosome. A man with hemophilia marries a normal woman whose father suffered from hemophilia. Determine the probability of having healthy children in this family.
51. A man with hemophilia marries a woman who does not have this disease. They have normal daughters and sons who marry non-hemophiliacs. Will hemophilia be discovered again in grandchildren, and what is the likelihood of patients appearing in the families of daughters and sons?
52. A proband man has a grandmother with normal vision and a color-blind grandfather on his mother’s side. This man's mother is colorblind, his father is normal. What are the genotypes of these individuals, and what will the proband’s children be like if he marries a woman genotypically similar to his sister?
53. Enamel hypoplasia (thin granular enamel, light brown teeth) is inherited as an X-linked dominant trait. In a family where both parents had this anomaly, a son was born with normal teeth. Determine the probability that their next child will also have normal teeth.
Linkage of features. Crossing over.
The number of genes in the human genotype is huge, and the number of chromosomes is 46 (23 pairs). Therefore, genes located on the same chromosome will be inherited together (linked), because will be transmitted from parents to children in the same linkage group. The number of linkage groups is equal to the haploid set of chromosomes. Linked inheritance in humans: hemophilia, color blindness, etc.
The linkage of genes in a chromosome can be complete or incomplete. With complete linkage, genes are always inherited together and do not result in crossing over. When linkage between allelic genes is incomplete, crossing over (exchange of sections of homologous chromosomes) is possible. When homologous regions are exchanged, linked genes diverge into different gametes:
The frequency of crossing over is directly proportional to the distance between genes, i.e. The farther the genes are from each other on a chromosome, the more often the exchange occurs; the closer they are located to each other, the less common the divergence of characteristics. Therefore, the segregation in the offspring when crossing diheterozygotes will be similar to Mendelian, but the ratio of hybrids will be different.
Tasks
54. The smooth shape of corn seeds dominates over wrinkled ones, and colored seeds dominate over uncolored ones. Both signs are linked. When crossing corn with smooth, colored seeds with a plant with wrinkled, uncolored seeds, the following offspring were obtained: colored smooth - 4152 individuals, colored wrinkled - 149, uncolored smooth - 152, uncolored wrinkled - 4163. Determine the distance between the genes.
55. In rats, dark coat color dominates over light coat color, pink color eye over red. Both signs are linked. In the laboratory, from crossing pink-eyed dark-haired rats with red-eyed light-haired rats, the following offspring were obtained: light red-eyed - 24, dark pink-eyed - 24, light pink-eyed - 26, dark red-eyed - 25. Determine the distance between the genes.
56. In humans, the Rh factor locus is linked to the locus that determines the shape of red blood cells and is located at a distance of 3 morganids from it (K. Stern, 1965). Rhesus positivity and elliptocytosis are determined by dominant autosomal genes. One of the spouses is heterozygous for both traits. At the same time, he inherited Rh-positiveness from one parent, elliptocytosis from the other. The second spouse is Rh negative and has normal red blood cells. Determine the percentages of probable genotypes and phenotypes of children in this family.
57. Cataract and polydactyly in humans are caused by dominant autosomal closely linked (i.e., not detecting crossing over) genes. However, not necessarily the genes for these anomalies may be linked, but also the cataract gene with the gene for normal hand structure and vice versa.
1. A woman inherited cataracts from her mother, and polydactyly from her father. Her husband is normal for both signs. What is more likely to be expected in children: the simultaneous appearance of cataracts and polydactyly, the absence of both of these signs, or the presence of only one anomaly—cataracts or polydactyly?
2. What kind of offspring can be expected in a family where the husband is normal and the wife is heterozygous for both characteristics, if it is known that the wife’s mother also suffered from both anomalies, and her father was normal?
58. Classic hemophilia and color blindness are inherited as recessive traits linked to the X chromosome. The distance between genes is determined to be 9.8 morganids.
1. A girl whose father suffers from both hemophilia and color blindness, and whose mother is healthy and comes from a family prosperous for these characteristics, marries a healthy man. Determine the likely phenotypes of the children from this marriage.
2. A woman whose mother suffered from color blindness and whose father suffered from hemophilia marries a man who suffers from both diseases. Determine the probability of children being born in this family simultaneously with both anomalies.
59. The color blindness gene and the night blindness gene, inherited through the X chromosome, are located at a distance of 50 morganids from each other (K. Stern, 1965). Both traits are recessive.
1. Determine the probability of having children simultaneously with both anomalies in a family where the wife has normal vision, but her mother suffered from night blindness, and her father suffered from color blindness, and the husband is normal in relation to both signs.
2. Determine the probability of having children simultaneously with both anomalies in a family where the wife is heterozygous for both characteristics and inherited both anomalies from her father, and the husband has both forms of blindness.
60. In humans, the genes for the Rh factor and elliptocytosis are located on the same chromosome at a distance of 3 morganids. Rh positivity and elliptocytosis are determined by dominant genes. The color blindness gene and the night blindness gene are located on the X chromosome at a distance of 50 morganids. Both traits are transmitted recessively.
1. A woman who is heterozygous for all characteristics, who has inherited all the abnormal genes from her mother, whose ancestors did not have crossing over, marries a man who suffers from both color and night blindness and is homozygous for both autosomal recessive genes. Determine the likely phenotypes of the children in this family.
2. A woman who is heterozygous for all characteristics marries an Rh-negative man who is normal in all other analyzed characteristics. It is known that the woman’s father was Rh-negative, had elliptocytosis, suffered from night blindness, but could distinguish colors normally. Determine the likely phenotypes of the children in this family.
