Sex-linked inheritance. Dihybrid sex-linkage tasks X-linked recessive gene for color blindness
Sex-linked traits are traits whose genes are located on the sex chromosomes. When solving problems, genes should be designated together with sex chromosomes, for example, X d - the gene for color blindness located on the X chromosome.
Problem 69. In humans, color blindness is caused by a recessive X-linked gene. The husband and wife have normal vision, despite the fact that their fathers are color blind. What is the probability of having color blind children in this family?
Solution: Let's designate the genes:
X D - normal color differentiation
X d - color blindness
According to the condition of the problem, the man has normal vision, therefore, his genotype is X D y (despite the fact that his father was color blind). A woman also has normal color vision, which means that her genotype must have a dominant normal X D gene. Since her father was color blind, she received the X chromosome from him with the color blindness gene and is a carrier of this gene: X D X d.
We write down the crossing scheme, write out all types of husband and wife gametes and all possible genotypes in F 1:
R O X D X d - O X D y
D X D X d X D y
F 1 X D X D, X D X d, X D y, X d y
Analyzing the F 1 offspring, we establish that ¼ of all offspring or ½ of all sons will be color blind, ¼ of all offspring or ½ of sons will have normal color vision. All girls will be healthy, but half of them are carriers of the color blindness gene.
Answer: the probability of having color-blind children in a given family is 25% of all offspring. By gender, only boys can be color blind, which means that the probability of being color blind is 50% of sons.
Problem 70. Hypoplasia of tooth enamel (darkening of tooth enamel) is determined by a dominant gene linked to the X chromosome. In a family where the husband and wife have tooth enamel hypoplasia, a son was born with normal teeth. What is the probability of the birth of children with this anomaly in this family?
Solution: Let's designate the genes:
X A - tooth enamel hypoplasia
Xa - normal teeth
According to the condition of the problem, the husband has tooth enamel hypoplasia, so his genotype is X A y. The wife also suffers from this anomaly, which means that her genotype contains a dominant gene that determines the X A disease. Since a son with normal teeth was born in this family, i.e. with the genotype X a y, and he received the lame X from his mother, which means that the mother's genotype must have a recessive gene X a. Thus, the genotype of the wife: X A X a.
R O X A X a - O X A y
G X A X a X A y
F 1 X A X A, X A X a, X A y, X a y - with normal teeth
with hypoplasia
After analyzing the F 1 offspring, we establish that all girls will be with hypoplasia (half of them are homozygous X A X A and half are heterozygous X A X a), half of the sons with hypoplasia (X A y) and half of the sons with normal teeth (X a y ). Thus, all the daughters and half of the sons will be with tooth enamel hypoplasia, i.e. ¾ (75%) of all offspring.
Answer: the probability of having children with hypoplasia of tooth enamel is 75%.
Problem 71. In humans, color blindness is caused by a recessive, X-linked gene, albinism is determined by an autosomal recessive gene. In a family of two normal spouses, a son was born with two anomalies. What is the probability of the next child being born healthy?
Solution: Let's designate the genes:
A - normal pigmentation
a - albinism
X D - normal color vision
X d - color blindness
Since both parents are normal in two ways, their genotype necessarily has a dominant gene from each allelic pair: A and X D. The son has both anomalies, hence his genotype: aaX d y. The child received one of the recessive genes for albinism from the mother, the other from the father, therefore, in the genotype of both the mother and the father, there must be a gene - a. So both parents are heterozygous for the first pair of genes: Aa. The son received the X chromosome with the gene for color blindness from the mother, which means that her genotype must contain the gene for color blindness. She is also heterozygous for the second pair: X D X d.
Thus, the genotypes of the parents: AaX D X d and AaX D y. For the analysis of offspring, it is most convenient to construct a Punnett lattice
R O AaX D X d - O AaX D y
G ACH D ACH d ACH D aC D
aX D aX d Ay ay
| : AH D: aX D: Ay: ay |
| AH D: AAH D X D : Aah D X D : AAH D y : Aah D y |
| aX D: Aah D X D: aaX D X D: Aah D y: aah y |
| AX d: AAH D X d : AaX D X d: AAX d y: AaX d y |
| aX d: AaX D X d: aaX D X d: AaX d y: aaX d y |
We emphasize those genotypes that correspond to normal phenotypes in both traits, i.e. genotypes in which there is at least one dominant gene from each pair (A and X D). We get ¾ among girls and 3/8 among boys, i.e. 9/16 from all offspring.
Answer: the probability of the birth of the next child without anomalies is 9/16 of all offspring, including 3/4 of girls and 3/8 of boys.
Problem 72. In humans, hemophilia (blood clotting disorder) is determined by a recessive gene linked to the X chromosome. A healthy woman whose father was a hemophiliac married a healthy man whose father also had hemophilia. What is the probability of the birth of healthy children and hemophiliacs in this family?
Problem 73. In some breeds of chickens, the striped color of the plumage is determined by the dominant gene linked to the X chromosome, and the white color is determined by the recessive gene, also localized on the X chromosome. In chickens, the homogametic sex is male, and the heterogametic sex is female. When crossing white hens with striped roosters, all offspring in F 1 turned out to be with striped plumage. What splitting by phenotype and genotype will be obtained in F 2 when crossing roosters and hens from F 1 with each other?
Problem 74. When crossing white hens with striped roosters, half of the total F 1 offspring (half of roosters and hens) turned out to be with striped plumage and half with white. Determine the genotypes of parents and offspring, if it is known that the genes that determine the white color and striped plumage color are localized on the X chromosome. The striped color gene dominates the white gene. Homogametic sex in chickens is male.
Problem 75. On a poultry farm, when striped hens were crossed with striped roosters, offspring of 798 striped hens, 400 striped hens and 397 white hens were obtained. Determine the genotypes of parents and offspring. The genes that determine white and striped plumage are linked to the X chromosome. The white gene is recessive to the striped gene. The heterogametic sex in chickens is female.
Problem 76. In Drosophila, the recessive white-eyed gene is linked to the X chromosome. The dominant gene allelic to it determines the red color of the eyes. Red-eyed females were crossed with white-eyed males. The entire generation of F 1 turned out to be red-eyed. What will be the eye color of the offspring of F 2 when males and females from F 1 are crossed with each other?
Problem 77. When red-eyed female fruit flies were crossed with red-eyed males, 86 red-eyed and females and 88 red-eyed and white-eyed males were found in the F 1 offspring. Determine the genotypes of the crossed individuals and offspring if it is known that the eye color genes are localized on the X chromosome and the red color gene dominates the white gene.
Problem 78. In cats, the genes that determine black and red coat color are located on the X chromosomes. None of these genes dominates, and if both genes are present in the genotype, a “three-haired” coloration is formed in the phenotype. What offspring can be expected from crossing a "three-haired" cat with a red cat? What is the probability of the appearance of "three-haired" cats from crossing a red cat with a black cat?
Problem 79. In humans, brown-eyedness is due to a dominant autosomal gene, hemophilia is recessive, linked to the X chromosome. A blue-eyed woman with normal blood clotting, whose father was a hemophilic, married a brown-eyed healthy man whose father was blue-eyed and hemophilic. What is the probability of having blue-eyed children with hemophilia in this family?
Problem 80. In humans, blue-eyedness is determined by a recessive autosomal gene, color blindness is determined by a recessive, linked to the X chromosome. A brown-eyed, normal-sighted woman whose mother was blue-eyed and whose father was color-blind married a blue-eyed, color-blind man. What is the probability of the birth of children with normal vision in this family and what will be their eye color?
Problem 81. In humans, right-handedness is due to a dominant autosomal gene, and hemophilia is recessive, localized on the X chromosome. A right-handed woman with normal blood clotting married a healthy right-handed man. They had a left-handed son suffering from hemophilia. Determine the probability of having healthy left-handed children in this family.
Problem 82. A left-handed woman with normal vision, whose father was color-blind, married a right-handed and color-blind man. It is known that the man's mother was left-handed. Determine the probability of birth of right-handed children with normal vision in this family. Left-handedness is determined by a recessive autosomal gene, and color blindness is determined by a recessive gene linked to the X chromosome.
Problem 83. In humans, albinism is caused by a recessive autosomal gene, and hypertrichosis (hairy ears) is caused by a gene located on the y chromosome. A man with hypertrichosis and normal pigmentation married a woman normal in both respects. They had a son - an albino with hypertrichosis. What is the probability of having children that are normal in both respects?
Problem 84. In humans, hypertrichosis (hairiness of the ears) is caused by a recessive gene linked to the Y chromosome, and color blindness is caused by a recessive gene linked to the X chromosome. A woman, normal in relation to both signs, whose father had both anomalies, married a man with normal vision and hypertrichosis. What is the prognosis for the offspring of this couple?
Problem 85. In humans, hypoplasia of tooth enamel (darkening of the teeth) is due to a dominant gene linked to the X chromosome, and albinism is due to a recessive autosomal gene. In a family where both spouses had normal pigmentation and suffered from tooth enamel hypoplasia, an albino son with normal teeth was born. What is the probability of birth in this family of children without anomalies?
Problem 86. In humans, hypoplasia of tooth enamel (darkening of the teeth) is determined by a dominant gene linked to the X chromosome, and color blindness is determined by a recessive gene, also localized on the X chromosome. A woman with normal teeth and normal vision, whose father was colorblind, married a man with hypoplasia and normal vision. What is the prognosis for offspring in this family?
Example. A woman suffering from color blindness married a man with normal vision. What will be the color perception of the sons and daughters of these parents?
Solution: in humans, color blindness (color blindness) is due to the recessive gene w, and normal color vision is due to its dominant allele W. The gene for color blindness is linked to the X chromosome. The Y chromosome does not have a corresponding locus and does not contain a gene that controls color vision. Therefore, in the diploid set of men, there is only one allele responsible for color perception, while in women there are two, since they have two X chromosomes.
The crossover scheme for this type of task is constructed somewhat differently from the schemes for traits not related to sex. These schemes indicate not only the symbols of genes, but also the symbols of the sex chromosomes of male and female individuals X and Y. This must be done to show the absence of the second allele in male and female individuals.
So, according to the condition of the problem, the crossover scheme will look like this. A woman has two X chromosomes and each has a recessive gene for color blindness (Xw). A man has only one X chromosome, which carries the gene for normal color vision (X W), and a Y chromosome that does not contain the allelic gene. First, we designate the two X chromosomes of a woman and indicate that they contain genes for color blindness (X w X w). Then we denote the male chromosomes (X W Y). Below we write out the gametes that they produce. In a woman, they are the same, they all contain the X chromosome with the gene for color blindness. In a man, half of the gametes carry the X chromosome with the gene for normal color perception, and half the Y chromosome that does not contain the allele. After that, determines the genotypes F 1 .
All girls receive one X chromosome from their father, which contains the gene for normal vision (W), and another X chromosome from their mother, which contains the gene for color blindness (w).
Thus, girls will have two Ww genes, and since the gene for normal vision dominates, they will not have color blindness. All boys receive their single X chromosome from their mother and the gene for color blindness (w) located in it. Since they do not have the second allele, they will suffer from color blindness.
58. The daughter of a color blind person marries the son of a color blind person, and the bride and groom can distinguish colors normally. What will their children's vision be like?
59. Men and women who are healthy in terms of color vision have a son suffering from color blindness, who has a healthy daughter. Determine the genotypes and phenotypes of parents, children and grandchildren.
60. Men and women who are healthy in terms of color vision have a healthy daughter who has one healthy son and one color-blind son. Determine the genotypes and phenotypes of parents, children and grandchildren.
61. Men and women who are healthy in terms of color vision have a healthy daughter who has five healthy sons. Determine the genotypes and phenotypes of parents, children and grandchildren.
62. What phenotypes can be expected for a Y-linked trait in the following descendants of a father with this trait:
a) sons
b) daughters
c) grandchildren from sons,
d) granddaughters from daughters,
e) granddaughters from sons,
f) grandchildren from daughters.
63. Hypertrichosis (growth of hair on the edge of the auricle) is inherited as a trait linked to the Y chromosome, which manifests itself only by the age of 17. Common ichthyosis (scaly and patchy thickening of the skin) is inherited as a recessive, X-linked trait. In a family where the woman is healthy in relation to both signs, and the husband suffers only from hypertrichosis, a boy was born with signs of ichthyosis. Determine: a) the probability of manifestation of hypertrichosis in this child; b) the probability of the birth of children in this family without both anomalies. What gender will they be?
64. In humans, the dominant gene P determines persistent rickets, which is inherited sex-linked. What is the probability of having sick children if the mother is heterozygous for the rickets gene, and the father is healthy?
65. The gene for eye color in Drosophila flies is located on the X chromosome. Red (normal) eyes (W) are dominant over white eyes (w). Determine the phenotype and genotype of the offspring F 1 and F 2 if you cross a white-eyed female with a red-eyed male.
66. In Drosophila flies, the gene for normal eye color (red) W dominates over white-eyed w, the gene for abnormal abdominal structure - over the gene for its normal structure. These pairs of genes are located on the X chromosome at a distance of three morganids. Determine the probability of different genotypes and phenotypes in the offspring from crossing a heterozygous female for both traits with a male with normal eye color and normal abdominal structure.
67. The H gene determines normal blood clotting in humans, and the h gene determines hemophilia. A woman heterozygous for the hemophilia gene married a man with normal blood clotting. Determine the phenotype and genotype of children that can be born from such a marriage. What is the type of trait inheritance?
68. In Plymouth Rock chickens, the dominant variegated gene is linked to the X chromosome. The recessive allele, the black gene, is observed in Austrolon chickens. Which cross will allow early sex marking of chicks?
1. A person has two types of blindness, and each is determined by its recessive autosomal gene, which are not linked. What is the probability of having a blind child if the father and mother suffer from the same type of blindness and both are dihomozygous? What is the probability of having a blind child if both parents are dihomozygous and suffer from different types hereditary blindness?
Explanation:
First cross:
R: AAvv x AAvv
G: Av x Av
F1: AABB is a blind child.
The law of uniformity appears. The probability of having a blind child is 100%.
Second cross:
R: AAvv x aaBB
G: Av x aV
F1: AaBv is a healthy child.
The law of uniformity appears. Both types of blindness are absent. The probability of having a blind child is 0%.
2. In humans, color blindness is caused by an X-linked recessive gene. Thalassemia is inherited as an autosomal dominant trait and occurs in two forms: in homozygotes it is severe, often fatal, in heterozygotes it is mild.
A woman with mild thalassemia and normal vision, married to a color-blind man who is healthy for the thalassemia gene, has a color-blind son with mild thalassemia. What is the probability of having this pair of children with both anomalies? Determine the genotypes and phenotypes of possible offspring.
Explanation:
R: AaXDXd x aaXdY
G: AXD, aXd, AXd, aXD x aXd, aY
F1: AaXdY - colorblind boy with mild thalassemia
AaXDXd - a girl with normal vision and mild thalassemia
aaXdXd - colorblind girl without thalassemia
AaXdXd - colorblind girl with mild thalassemia
aaXDХd - a girl with normal vision without thalassemia
AaXDY - boy with normal vision and mild thalassemia
aaXdY - colorblind boy without thalassemia
aaXDY - boy with normal vision and no thalassemia
That is, eight variants of the genotype are obtained with an equal probability of occurrence. The probability of having a child with a mild form of thalassemia and color blindness is 2/8 or 25% (12.5% chance of having a boy and 12.5% of having a girl). The probability of having a color-blind child with a severe form of thalassemia is 0%.
3. A blue-eyed fair-haired man and a diheterozygous brown-eyed dark-haired woman entered into marriage. Determine genotypes married couple, as well as possible genotypes and phenotypes of children. Set the probability of having a child with a dihomozygous genotype.
Explanation: A - Brown eyes
a blue eyes
B - dark hair
c - blond hair
R: aavv x AaVv
G: av x AB, av, av, aB
F1: AaBv - brown eyes, dark hair
aavv - blue eyes, blonde hair
Aaww - brown eyes, blond hair
aaBv - blue eyes, dark hair
The probability of having a child with each of the genotypes is 25%. (and the probability of having a child with a dihomozygous genotype (aavb) - 25%)
The signs are not sex-linked. This is where the law of independent inheritance comes into play.
4. When a gray (a) shaggy rabbit was crossed with a black shaggy rabbit, splitting was observed in the offspring: black shaggy rabbits and gray shaggy ones. In the second crossing of phenotypically the same rabbits, offspring were obtained: black shaggy rabbits, black smooth-haired, gray shaggy, gray smooth-haired. What law of heredity is manifested in these crosses?
Explanation:
A - black color
a - gray color
B - furry rabbit
c - smooth-haired rabbit
First cross:
R: aaBB x AaBB
F1: AaBB - black fluffy rabbits
aaBB - gray furry rabbits
Second cross:
R: aaBv x AaBv
G: аВ, АВ x АВ, av, АВ, аВ
F1: 8 genotypes and 4 phenotypes
АаВВ, 2АаВв - gray furry rabbits
Aavv - black smooth-haired rabbits
ааВВ, ааВв - gray furry rabbits
aavv - gray smooth-haired rabbits
In this case, the law of independent inheritance applies, since the presented signs are inherited independently.
5. For a crested (A) green (B) female, an analyzing cross was carried out, four phenotypic classes were obtained in the offspring. The resulting crested offspring were crossed among themselves. Is it possible to get offspring without a crest in this crossing? If so, what gender will it be, what phenotype? In canaries, the presence of a crest depends on an autosomal gene, the color of plumage (green or brown) depends on a gene linked to the X chromosome. The heterogametic sex in birds is female.
Explanation:
First cross:
R: AaHVU x aaHvHv
G: AHV, AHV, AU, AU X AHV
F1: АаХВХв - crested green male
ааХВХв - green male without crest
AaHvU - crested brown female
We cross a male and a female with a crest:
R: AaHVHv x AaHvU
G: AHV, AHV, AHV, AHV x AHV, AU, AHV, AU
F2: we get 16 genotypes, among which only 4 phenotypes can be distinguished.
Phenotypes of individuals without crest:
Females: ааХВУ - green female without crest
ааХвУ - brown female without crest
Males: ааХВХв - green male without crest
ааХвХв - brown male without crest.
6. In crossing female Drosophila with normal wings and normal eyes and males with reduced wings and small eyes, all offspring had normal wings and normal eyes. The females obtained in the first generation were backcrossed with the original parent. The shape of the wings in Drosophila is determined by an autosomal gene, the gene for eye size is located on the X chromosome. Make crossbreeding schemes, determine the genotypes and phenotypes of parental individuals and offspring in crosses. What are the laws of crossbreeding?
Explanation:
A - normal wings
a - reduced wings
XB - normal eyes
First cross:
R: AAHVHV x vvHvU
G: AHV x AHV, AU
AaHVHv - normal wings, normal eyes
AaHVU - normal wings, normal eyes
Second cross:
R: AaHVHv x aaHvN
G: AHV, AHV, AHV, AHV x AHV, AU
АаХВХв, АаХВУ - normal wings, normal eyes
ааХвХв, ааХвУ - reduced wings, small eyes
AaHvHv, AaHvU - normal wings, small eyes
ааХВХв, ааХВУ - reduced wings, normal eyes
The law of sex-linked inheritance applies here (the eye shape gene is inherited with the X chromosome), and the wing gene is inherited independently.
7. When crossing a Drosophila fly with a gray body (A) and normal wings (B), with a fly with a black body and twisted wings, 58 flies were obtained with a gray body and normal wings, 52 with a black body and twisted wings, 15 - with a gray body and swirling wings, 14 - with a black body and normal wings. Make a scheme for solving the problem. Determine the genotypes of the parent individuals, offspring. Explain the formation of four phenotypic classes. What law applies in this case?
Explanation: A - gray body
a black body
B - normal wings
c - swirling wings
Crossing: R: AaVv x aavv
G: AB, av, av, aB x av
F1: AaBv - gray body, normal wings - 58
aavv - black body, swirling wings - 52
Aavv - gray body, swirling wings - 15
aaBv - black body, normal wings - 14
Genes A and B and a and b are linked, so they form groups of 58 and 52 individuals, and in the case of the remaining two groups, crossing over occurred and these genes ceased to be linked, and therefore formed 14 and 15 individuals.
8. When analyzing the crossing of a diheterozygous tall tomato plant with round fruits, a splitting of offspring according to the phenotype was obtained: 38 tall plants with round fruits, 10 tall plants with pear-shaped fruits, 10 dwarf plants with round fruits, 42 dwarf plants with pear-shaped fruits. Make a crossbreeding scheme, determine the genotypes and phenotypes of the original individuals, offspring. Explain the formation of four phenotypic classes.
Explanation:
A is a tall plant
a - dwarf plant
B - round fruits
c - pear-shaped fruits
R: AaBv x aavb
G: AB, av, aB, Av x av
F1: AaBv - tall plants with round fruits - 38
aavv - dwarf plants with pear-shaped fruits - 42
aaBv - dwarf plants with round fruits - 10
Aavv - tall plants with pear-shaped fruits - 10
Here we can distinguish two groups of signs:
1. AaBv and aavb - in the first case, A and B are inherited linked, and in the second - a and c.
2. aaBv and Aavb - there was a crossing over.
9. In humans, non-red hair is dominant over red hair. Father and mother are heterozygous redheads. They have eight children. How many of them can be red? Is there a definite answer to this question?
Explanation: A - red hair
a red hair
R: Aa x Aa
G: A, a x A, a
F1: AA: 2Aa: aa
Splitting by genotype - 1:2:1.
Splitting by phenotype - 3:1. Therefore, the probability of having a non-red child is 75%. The probability of having a red-haired child is 25%.
There is no unequivocal answer to the question, since it is impossible to assume the genotype of the unborn child, since sex cells with different genotypes can be found.
10. Determine the genotypes of parents in a family where all sons are color blind and daughters are healthy.
Explanation: XDXd - healthy girl
XdY - the boy is colorblind
This situation will be more possible if the mother is color blind (since the female sex is homogametic), and the father is healthy (heterogametic sex).
Let's write a crossover scheme.
P: XdXd x XDY
G: Xd x XD, Y
F1: XDXd - a healthy girl, but a carrier of the gene for color blindness.
XdY - colorblind boy
11. In humans, glaucoma is inherited as an autosomal recessive trait (a), and Marfan's syndrome, accompanied by an anomaly in the development of connective tissue, is inherited as an autosomal dominant trait (B). The genes are located in different pairs of autosomes. One of the spouses suffers from glaucoma and had no ancestors with Marfan's syndrome, and the second is diheterozygous according to these characteristics. Determine the genotypes of the parents, the possible genotypes and phenotypes of the children, the probability of having a healthy child. Make a scheme for solving the problem. What law of heredity is manifested in this case?
Explanation: glaucoma is a recessive trait and manifests itself only with homozygote, and Marfan syndrome manifests itself both with hetero- and homozygote, but is a dominant trait, respectively, we will determine the genotypes of the parents: one parent suffers from glaucoma - aa, but does not suffer from Marfan syndrome - cc, and the second parent is heterozygous for both traits - AaBv.
R: aavv x AaVv
G: av x AB, av, av, aB
F1: AaBv - normal vision + Marfan's syndrome
aavv - glaucoma
Aavv - normal vision, no Marfan syndrome - healthy child
aaBv - glaucoma + Marfan's syndrome
By drawing the Punnett lattice, you can see that the probability of having each child is the same - 25%, which means that the probability of having a healthy child will be the same.
The genes of these traits are not linked, which means that the law of independent inheritance is manifested.
12. Undersized (dwarf) tomato plants with ribbed fruits and plants of normal height with smooth fruits were crossed. In the offspring, two phenotypic groups of plants were obtained: undersized and smooth fruits and normal height with smooth fruits. When crossing undersized tomato plants with ribbed fruits with plants having a normal stem height and ribbed fruits, all offspring had a normal stem height and ribbed fruits. Create crossover patterns. Determine the genotypes of parents and offspring of tomato plants in two crosses. What law of heredity is manifested in this case?
Explanation: in the first crossing, a dihomozygote is crossed with a homozygous plant for one trait and a heterozygous plant for another (to understand this, you need to write several options, this offspring is obtained only with such parents). in the second crossing, everything is simpler - two dihomozygotes are crossed (only the second parent will have one dominant trait).
a - undersized individuals
A - normal height
c - ribbed fruits
B - smooth fruits
P: aavb x aabb
F1: aaBb - undersized individuals with smooth fruits
AaBv - normal height, smooth fruits
P: aavv x AAiv
F1: Aavv - normal height, smooth fruits.
In both cases, the law of independent inheritance is manifested, since these two traits are inherited independently.
13. Based on the pedigree shown in the figure, determine and explain the nature of the inheritance of the trait highlighted in black. Determine the genotypes of parents, offspring, indicated on the diagram by the numbers 2, 3, 8, and explain their formation.
Explanation: since in the first generation we see uniformity, and in the second generation - splitting 1: 1, we conclude that both parents were homozygous, but one for a recessive trait, and the other for a dominant one. That is, in the first generation, all children are heterozygous. 2 - Aa, 3 - Aa, 8 - aa.
14. When crossing a motley crested (B) chicken with the same rooster, eight chickens were obtained: four motley crested chickens, two white (a) crested and two black crested. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, explain the nature of the inheritance of traits and the appearance of individuals with variegated coloration. What laws of heredity are manifested in this case?
Explanation: such splitting is possible only if the parents are heterozygous in color, that is, the variegated color has the genotype - Aa
AA - black color
aa - white color
Aa - variegated coloring
P: AaBB x AaBB
G: AB, aB
F1: AaBB - pied crested (4 chicks)
aaBB - white crested (two chickens)
AABB - black crested
In terms of color, the splitting by genotype and phenotype is the same: 1:2:1, since there is a phenomenon of incomplete dominance (an intermediate variant appears between both black and white), the traits are inherited independently.
15. In humans, the normal hearing gene (B) dominates the deafness gene and is located in the autosome; the gene for color blindness (color blindness - d) is recessive and linked to the X chromosome. In a family where the mother suffered from deafness, but had normal color vision, and the father had normal hearing (homozygous), colorblind, a colorblind girl with normal hearing was born. Make a scheme for solving the problem. Determine the genotypes of parents, daughters, possible genotypes of children and the probability of future birth in this family of color-blind children with normal hearing and deaf children.
Explanation: It can be seen from the conditions of the problem that the mother is heterozygous for the deafness gene and homozygous for the blindness gene, and the father has the blindness gene and is heterozygous for the deafness gene. Then the daughter will be homozygous for the blindness gene and heterozygous for the deafness gene.
P: (mother)XDXd x (father)XdYBB
daughter - XdXdBb - colorblind, normal hearing
Gametes - XDb, Xdb, XdB, YB
Children: XDXdBb - normal vision, normal hearing
XDYBb - normal vision, normal hearing
XdXdBb - colorblind, normal hearing
XdYBb - colorblind, normal hearing
Splitting: 1:1:1:1, that is, the probability of a color-blind child with normal hearing is 50%, and the probability of a color-blind child being born is 0%.
16. The husband and wife have normal vision, despite the fact that the fathers of both spouses suffer from color blindness (color blindness). The gene for color blindness is recessive and linked to the X chromosome. Determine the genotypes of husband and wife. Make a scheme for solving the problem. What is the probability of having a son with normal vision, a daughter with normal vision, a color blind son, a color blind daughter?
Explanation: Let's say the mothers of the husband and wife were healthy.
We will also write down the possible genotypes of the parents of the husband and wife.
P: XDXD x XdY XDXD x XdY
↓ ↓
XDXd x XDY
↓
Possible genotypes of children:
XDXD - healthy girl
XDY - healthy boy
XDXd - healthy girl
XdY - colorblind boy
The probability of having a child with each of the genotypes is 25%. The probability of having a healthy girl is 50% (in one case, the child is heterozygous, in the other, homozygous). The probability of having a colorblind girl is 0%. The probability of having a color-blind boy is 25%.
17. In sowing peas, the yellow color of the seeds dominates over the green, the convex shape of the fruit - over the fruits with a constriction. When a plant with yellow convex fruits was crossed with a plant with yellow seeds and fruits with constriction, 63 plants with yellow seeds and convex fruits were obtained. 58 with yellow seeds and constricted fruits, 18 with green seeds and convex fruits, and 20 with green seeds and constricted fruits. Make a scheme for solving the problem. Determine the genotypes of the original plants and descendants. Explain the emergence of different phenotypic groups.
Explanation:
A - yellow color
a green color
B - convex shape
c - fruits with constriction
After carefully reading the condition of the problem, one can understand that one parent plant is diheterozygous, and the second is homozygous for the shape of the fruit, and heterozygous for the color of the seed.
Let's write a scheme for solving the problem:
P: AaBv x Aavb
G: AB, av, av, aB x av, av
F1: This results in a 3:1 split and the following first-generation offspring:
63 - A_Vv - yellow seeds, convex fruits
58 - A_vv - yellow seeds, fruits with constriction
18 - aaBv - green seeds, convex shape of the fruit
20 - aavv - green seeds, fruits with constriction
Here we observe the law of independent inheritance, since each trait is inherited independently.
18. In snapdragons, the red color of flowers incompletely dominates over white, and narrow leaves over wide ones. Genes are located on different chromosomes. Plants with pink flowers and leaves of intermediate width are crossed with plants with white flowers and narrow leaves. Make a scheme for solving the problem. What offspring and in what ratio can be expected from this crossing? Determine the type of crossing, genotypes of parents and offspring. What is the law in this case.
Explanation: AA - red color
Aa - pink coloring
aa - white color
BB - narrow leaves
Вв - leaves of intermediate width
cc - wide leaves
Crossbreeding:
R: AaVv x aaVV
G: AB, av, AV, aB x aB
F1: AaBB - pink flowers, narrow leaves
aaBv - white flowers, leaves of intermediate width
AaBv - pink flowers, leaves of intermediate width
aaBB - white flowers, narrow leaves
The probability of flowers appearing with each of the genotypes is 25%.
Crossing is dihybrid (since the analysis is based on two traits).
In this case, the laws of incomplete dominance and independent inheritance of traits apply.
Tasks for independent solution
1. In dogs, black coat dominates over brown, and long coat over short (genes are not linked). Offspring were obtained from a black long-haired female during analyzing crosses: 3 black long-haired puppies, 3 brown long-haired puppies. Determine the genotypes of parents and offspring corresponding to their phenotypes. Make a scheme for solving the problem. Explain your results.
2. In sheep, gray color (A) of wool dominates over black, and horn (B) dominates over polled (hornless). The genes are not linked. In the homozygous state, the gray color gene causes the death of embryos. What viable offspring (by phenotype and genotype) and in what ratio can be expected from crossing a diheterozygous sheep with a heterozygous gray polled male? Make a scheme for solving the hadachi. Explain your results. What law of heredity is manifested in this case?
3. In maize, the recessive gene "shortened internodes" (b) is on the same chromosome as the recessive gene "rudimentary panicle" (v). When conducting an analyzing cross of a diheterozygous plant with normal internodes and a normal panicle, offspring were obtained: 48% with normal internodes and a normal panicle, 48% with shortened internodes and a rudimentary panicle, 2% with normal internodes and a rudimentary panicle, 2% with shortened internodes and normal panicle. Determine the genotypes of parents and offspring. Make a scheme for solving the problem. Explain your results. What law of heredity is manifested in this case?
4. When crossing a corn plant with smooth colored seeds with a plant that produces wrinkled uncolored seeds (genes are linked), the offspring turned out to be with smooth colored seeds. When analyzing the crossing of hybrids from F1, plants were obtained with smooth colored seeds, wrinkled uncolored seeds, wrinkled colored seeds, and smooth uncolored seeds. Make a scheme for solving the problem. Determine the genotypes of the parents, offspring F1 and F2. What laws of heredity are manifested in these crosses? Explain the appearance of four phenotypic groups of individuals in F2?
5. When a corn plant with smooth colored seeds was crossed with a plant with wrinkled uncolored seeds (genes linked), the offspring turned out to have smooth colored seeds. With further analyzing crossing of the hybrid from F1, plants with seeds were obtained: 7115 with smooth colored seeds, 7327 with wrinkled uncolored seeds, 218 with wrinkled colored seeds, 289 with smooth uncolored seeds. Make a scheme for solving the problem. Determine the genotypes of parents, offspring F1, F2. What law of heredity is manifested in F2? Explain what your answer is based on.
6. In humans, cataracts (eye disease) depend on a dominant autosomal gene, and ichthyosis (skin disease) depends on an X-linked recessive gene. Woman with healthy eyes and with normal skin, whose father suffered from ichthyosis, marries a man suffering from cataracts and healthy skin whose father did not have these diseases. Make a scheme for solving the problem. Determine the genotypes of the parents, the possible genotypes and phenotypes of the children. What laws of heredity are manifested in this case?
7. When crossing white rabbits with shaggy hair and black rabbits with smooth hair, offspring were obtained: 50% black shaggy and 50% black smooth. When crossing another pair of white rabbits with shaggy hair and black rabbits with smooth hair, 50% of the offspring turned out to be black shaggy and 50% white shaggy. Make a diagram of each cross. Determine the genotypes of parents and offspring. Explain what law is manifested in this case?
8. When a watermelon plant with long striped fruits was crossed with a plant with round green fruits, plants with long green and round green fruits were obtained in the offspring. When the same watermelon with long striped fruits was crossed with a plant with round striped fruits, all offspring had round striped fruits. Make a diagram of each cross. Determine the genotypes of parents and offspring. What is the name of this crossing and why is it carried out?
9. A dark-haired, blue-eyed woman, dihomozygous, married a dark-haired, blue-eyed man, heterozygous for the first allele. Dark color hair and brown eyes are dominant traits. Determine the genotypes of parents and offspring, types of gametes and probable genotypes of children.
10. A dark-haired woman with curly hair, heterozygous for the first trait, married a man with dark smooth hair, heterozygous for the first allele. Dark and curly hair are dominant traits. Determine the genotypes of the parents, the types of gametes they produce, the probable genotypes and phenotypes of the offspring.
11. A dark-haired, brown-eyed woman, heterozygous for the first allele, married a fair-haired, brown-eyed man, heterozygous for the second trait. Dark hair and brown eyes are dominant traits, blond hair and blue eyes are recessive traits. Determine the genotypes of the parents and the gametes they produce, the probable genotypes and phenotypes of the offspring.
12. A red-eyed gray (A) Drosophila, heterozygous for two alleles, was crossed with a red-eyed black (XB) Drosophila, heterozygous for the first allele. Determine the genotypes of the parents, the gametes they produce, the numerical ratio of the splitting of offspring by genotype and phenotype.
13. A black hairy rabbit heterozygous for two alleles was crossed with a white hairy rabbit heterozygous for the second allele. Black shaggy fur is dominant traits, white smooth fur is recessive traits. Determine the genotypes of the parents and the gametes they produce, the numerical ratio of the splitting of the offspring by phenotype.
14. The mother has the 3rd blood group and positive Rh factor, and the father has the 4th blood group and the Rh factor is negative. Determine the genotypes of the parents, the gametes they produce, and the possible genotypes of the children.
15. From a black cat, one tortoiseshell and several black kittens were born. These traits are sex-linked, that is, color genes are found only on the sex X chromosomes. The gene for black color and the gene for red color gives incomplete dominance, when these two genes are combined, a tortie color is obtained. Determine the genotype and phenotype of the father, the gametes that the parents produce, the sex of the kittens.
16. A heterozygous gray female Drosophila was crossed with a gray male. These traits are sex-linked, that is, the genes are found only on the sex X chromosomes. The gray color of the body dominates over the yellow. Determine the genotypes of the parents, gametes. which they produce, and the numerical splitting of offspring by sex and body color.
17. In tomato, the genes that cause high plant growth (A) and the round shape of the fruit (B) are linked and localized on the same chromosome, and the genes that cause low growth and pear-shaped shape are in the autosome. They crossed a heterozygous tomato plant, which has a high growth and a round fruit shape, with a low pear-bearing plant. Determine the genotypes and phenotypes of the offspring of the parents, gametes formed in meiosis if there was no chromosome crossing.
18. In Drosophila, the dominant genes for normal wing and gray body color are linked and localized on one chromosome, and the recessive genes for rudimentary wing and black body color are on the other homologous chromosome. We crossed two diheterozygous Drosophila with normal wings and a gray body color. Determine the genotype of the parents and gametes formed without chromosome crossing, as well as the numerical ratio of the splitting of the offspring by genotype and phenotype.
19. What are the genotypes of parents and children if a fair-haired mother and a dark-haired father had five children in marriage, all dark-haired? What is the law of heredity?
20. What are the genotypes of parents and offspring, if all offspring are black from crossing a cow with a red coat color with a black bull? Determine the dominant and recessive genes and the nature of dominance.
21. What phenotypes and genotypes are possible in children if the mother has the first blood type and homozygous Rh-positive factor, and the father has the fourth blood group and Rh negative factor(recessive trait)? Determine the probability of having children with each of these characteristics.
22. A blue-eyed child was born in the family, similar in this feature to his father. The child's mother is brown-eyed, the grandmother maternal line- blue-eyed, and grandfather - brown-eyed. Paternal grandparents are brown-eyed. Determine the genotypes of the parents and paternal grandparents. What is the probability of having a brown-eyed child in this family?
23. Woman with blonde hair and with a straight nose, married a man with dark hair and a Roman nose, heterozygous for the first trait and homozygous for the second. Dark hair and a Roman nose are dominant traits. What are the genotypes and gametes of the parents? What are the likely genotypes and phenotypes of the children?
24. Several kittens were born from a tortoiseshell cat, one of which turned out to be a red cat. In cats, coat color genes are sex-linked and are found only on the X chromosome. Tortoiseshell coat color is possible with a combination of the gene for black and red color. Determine the genotypes of the parents and the phenotype of the father, as well as the genotypes of the offspring.
25. A heterozygous gray female Drosophila was crossed with a gray male. Determine the gametes produced by the parents, as well as the numerical ratio of the splitting of hybrids by phenotype (by sex and body color) and genotype. These traits are sex-linked and are found only on the X chromosomes. Gray body color is a dominant trait.
26. In maize, the dominant genes for brown color and smooth seed shape are linked and localized on one chromosome, and the recessive genes for white color and wrinkled shape are on the other homologous chromosome. What offspring according to the genotype and phenotype should be expected when a diheterozygous plant with white smooth seeds is crossed with a plant with white wrinkled seeds. Crossing over did not occur during meiosis. Determine the gametes produced by the parents.
27. In corn, the dominant genes for brown color and smooth seed shape are linked and localized on one chromosome, and the recessive genes for white color and wrinkled shape are on the other homologous chromosome. What offspring according to the genotype and phenotype should be expected when crossing a diheterozygous plant with white smooth seeds with a homozygous plant with dark smooth seeds. In meiosis, crossing over occurs. Determine the gametes produced by the parents without crossing over and after crossing over.
28. When a shaggy white rabbit was crossed with a shaggy black rabbit, one smooth white rabbit appeared in the offspring. Determine the genotypes of the parents. In what numerical ratio can we expect the splitting of the offspring according to the genotype and phenotype?
29. A hunter bought a dog that has short hair. It is important for him to know that she is purebred. What actions will help the hunter to determine that his dog does not carry recessive genes - long hair? Make a scheme for solving the problem and determine the ratio of the genotypes of the offspring obtained from crossing a purebred dog with a heterozygous one.
30. A man suffers from hemophilia. His wife's parents are healthy on this basis. The hemophilia gene (h) is located on the sex X chromosome. Make a scheme for solving the problem. Determine the genotypes of a married couple, possible offspring, the probability of having daughters who are carriers of this disease.
31. Hypertrichosis is transmitted in a person with a Y chromosome, and polydactyly (multi-fingeredness) is an autosomal dominant trait. In a family where the father had hypertrichosis and the mother had polydactyly, a normal daughter was born. Make a scheme for solving the problem and determine the genotype of the born daughter and the probability that the next child will have two abnormal features.
32. Diheterozygous tomato plants with rounded fruits (A) and with pubescent leaves (B) were crossed with plants having oval fruits and non-pubescent leaf epidermis. The genes responsible for the structure of the leaf epidermis and the shape of the fruit are inherited linked. Make a scheme for solving the problem. Determine the genotypes of the parents, the genotypes and phenotypes of the offspring, the probability of the appearance of plants with recessive traits in the offspring.
33. When crossing a tomato with a purple stem (A) and red fruits (B) and a tomato with a green stem and red fruits, 750 plants with a purple stem and red fruits and 250 plants with a purple stem and yellow fruits were obtained. The dominant genes for purple stem color and red fruit color are inherited independently. Make a scheme for solving the problem. Determine the genotypes of the parents, offspring in the first generation and the ratio of genotypes and phenotypes in the offspring.
34. A Datura plant with purple flowers (A) and smooth boxes (c) was crossed with a plant with purple flowers and spiny boxes. The following phenotypes were obtained in the offspring: with purple flowers and spiny boxes, with purple flowers and smooth boxes, with white flowers and spiny boxes, with white flowers and smooth boxes. Make a scheme for solving the problem. Determine the genotypes of parents, offspring and the possible ratio of phenotypes. Establish the nature of inheritance of traits.
35. Two snapdragon plants with red and white flowers were crossed. Their offspring turned out to be with pink flowers. Determine the genotypes of parents, hybrids of the first generation and the type of inheritance of traits.
36. A brown (a) long-haired (c) female is crossed with a homozygous black (A) short-haired (B) male (genes are not linked). Draw up a scheme for solving the problem and determine the genotypes and phenotype ratio of the offspring of their first generation. What is the ratio of genotypes and phenotypes of the second generation from crossing diheterozygotes. What genetic patterns are manifested in this crossing?
37. In pigs, black bristle color (A) dominates over red, long bristle (B) - over short (genes are not linked). A diheterozygous black with long stubble was crossed with a homozygous black with short stubble female. Make a scheme for solving the problem. Determine the genotypes of parents, offspring, offspring phenotypes and their ratio.
38. The absence of small molars in humans is inherited as a dominant autosomal trait. Determine the genotypes and phenotypes of parents and offspring if one of the spouses has small molars, and the other is heterozygous for this gene. Make a scheme for solving the problem and determine the probability of having children who do not have small molars.
When solving such problems, one should use the principles of solving problems for dihybrid crossing, taking into account the features inheritance of sex-linked traits.
Task 7-14
Recessive genes encoding signs of hemophilia and color blindness are linked to the X chromosome. A man with hemophilia marries healthy woman, whose father was colorblind, but not hemophilic. What offspring will come from the marriage of their daughter with a healthy man?
A - normal clotting, a - hemophilia,
B - normal color perception, b - color blindness.
- The genotype of a man is X aB Y, since he carries a sign of hemophilia and is not color blind.
- The woman's father was colorblind, hence she received the recessive gene for colorblindness from him. The second allele of this gene is in a dominant state, since the woman is healthy. On the basis of hemophilia, the woman is homozygous, as she is healthy (dominant trait), and her father was healthy. The genotype of a woman is X AB X Ab.
- The daughter's husband's genotype is X AB Y, since he does not suffer from either color blindness or hemophilia.
- On the basis of hemophilia, the daughter is heterozygous, since she can only receive a recessive gene from her father, and only a dominant gene from a homozygous mother. Her father gave her a dominant gene for color blindness, and her mother could have passed her both a dominant and a recessive gene. Therefore, the daughter's genotype may be X aB X Ab or X aB X AB. The problem has two solutions.
In the first case, 25% of children (half of the boys) will suffer from hemophilia, in the second, half of the boys will suffer from hemophilia, and half will be color blind.
Task 7-15
The recessive genes for hemophilia and color blindness are associated with the X chromosome. What offspring will be obtained from the marriage of a man with hemophilia and a woman with color blindness (homozygous for the absence of hemophilia)?
Task 7-16
A man suffering from hemophilia and color blindness married a healthy woman who did not carry the genes for these diseases. What is the probability that a child from the marriage of his daughter with a healthy man:
- there will be one of these diseases;
- Will there be both anomalies?
There is no crossing over between the genes for color blindness and hemophilia.
Task 7-17
The recessive genes that determine the development of hemophilia and color blindness can be located on the human X chromosome. A woman has a father with hemophilia but not color blindness and a healthy hemophiliac (homozygous) mother who is colorblind. This woman is marrying a healthy man. What is the probability of giving birth to a child with one anomaly, assuming that there is no crossing over between the genes for hemophilia and color blindness?
Read also other topics chapter VII "Inheritance of genes localized in sex chromosomes".
The gene for color blindness is linked to the X chromosome. normal in relation to vision, a man and a woman have a color-blind son and a healthy daughter. What will be the grandchildrenIf their children will marry healthy people?
Please solve the problems, it is very necessary, what, what, but biology is not mine! (1. The recessive gene for color blindness is located on the X chromosome, motherthe girl suffers from color blindness, and the father, like all his ancestors, distinguishes colors normally, the girl marries a healthy young man. person. The probability of being born healthy and suffering from color blindness
2. The recessive gene is on the X chromosome, the girl's father is healthy, her mother is healthy, but her father was ill with hemophilia (mother's father), the girl marries a healthy young man. What can you say about their future children and grandchildren?
Color blindness (color blindness) is inherited as a recessive trait controlled by a gene linked to the X chromosome. In a family where the wifeis a carrier of the gene for color blindness, and her husband suffers from color blindness. What is the probability that this family will have children with normal color discrimination?
1.
A color-blind man (color blind, a sex-linked trait) married a woman with normal vision, but who had a color-blind father. Can theyto be born a colorblind daughter? What is the probability of having the first two colorblind sons?
In chickens, silver plumage is a dominant trait. Silver plumage gene S (Silver) is localized in Z-chromosome. A silver rooster was crossed with a black hen. What are the possible consequences of this crossing?
Enamel hypoplasia is inherited as linked to X-chromosome dominant trait. In a family where both parents suffered from the marked anomaly, a son was born with normal teeth. Determine the probability that the next child will also have normal teeth.
One form of hemophilia is inherited as a recessive trait linked to X-chromosome. A man with hemophilia marries normal woman(whose father and mother are healthy). Will children suffer from this disease?
* A tortoiseshell (spotted) cat was crossed with a red cat. How will the splitting of hybrids by genotype and phenotype (by coat color and sex) go?
*Hypertension in humans is determined by a dominant autosomal gene, optic atrophy is caused by a sex-linked recessive gene. A woman with optic atrophy marries a man with hypertension whose father also had hypertension and whose mother was healthy. What is the probability that a child in this family will suffer from both anomalies (in %)? What is the probability of having a healthy baby (in %)?
* What are the genotypes of the parents, if a blue-eyed hemophilic boy was born from the marriage of a brown-eyed healthy (according to blood clotting) woman with a blue-eyed healthy man. Hemophilia and blue eyes are recessive traits, but the gene blue eyes located on the autosome, and the hemophilia gene is linked to X-chromosome.
A man suffering from color blindness and deafness married a woman with good hearing and vision. They had a son with deafness and normal vision and a daughtercolorblind with good hearing. determine the genotypes of parents and children, if both signs (deafness and color blindness) are recessive, deafness is an autosomal trait, and color blindness is linked to the x chromosome